Course Theory
Book 1. Foundations of Olympiad Geometry
Book 1. Foundations of Olympiad Geometry
- 1. Angles, Lines and Parallel Lines
- 2. Triangles I: Congruence
- 3. Triangles II: Similarity
- 4. Quadrilaterals
- 5. Circles I: Basic Circle Geometry
- 6. Areas I
- 7. Basic Constructions and Auxiliary Lines
- 8. Mixed Problems I
Chapter
Angles, Lines and Parallel Lines
This module teaches students to work confidently with vertical and adjacent angles, parallel lines, the angle sum of a triangle, exterior angles, and the first angle-chasing problems.
Key Idea
In angle problems, the main goal is not to guess from the diagram but to build a chain of equalities. One known angle often gives the next one through an adjacent angle, a vertical angle, parallel lines, the angle sum of a triangle, or an exterior angle.
This method is called angle chasing: we record known angles, transfer them along parallel lines, and gradually reach the required angle or the required parallelism.
Basic Facts
Adjacent angles sum to \(180^\circ\). Vertical angles are equal. If two lines are cut by a transversal, then for parallel lines alternate interior angles are equal, corresponding angles are equal, and same-side interior angles sum to \(180^\circ\).
The converses are also true: if alternate interior angles or corresponding angles are equal, then the two lines are parallel. The angles of a triangle sum to \(180^\circ\), and an exterior angle of a triangle equals the sum of the two non-adjacent interior angles.
When to Use This Method
Use angle chasing when the condition contains parallel lines, angle bisectors, exterior angles, extensions of sides, isosceles triangles, or asks you to prove parallelism.
If you need to prove that two lines are parallel, look for equal alternate interior or corresponding angles. If you need to find an angle in a triangle, start with the angle sum and exterior angles.
How to Recognise the Method
Typical signs are a transversal, a zigzag of several lines, an angle that can be moved to another place, or given ratios of angles. It is often useful to write all known angles directly on the diagram.
If an angle lies on a straight line, check its adjacent angle. If two lines look parallel, try to find a pair of alternate interior angles.
Typical Mistakes
Do not assume that lines are parallel just because they look parallel in the diagram. Do not transfer angles unless you have parallel lines or vertical angles.
A common mistake is to confuse an exterior angle with its adjacent interior angle. The exterior angle equals the sum of the two remote interior angles, not the whole angle sum of the triangle.
Mini-Checklist
1. Mark all given angles. 2. Use adjacent and vertical angles. 3. Transfer angles along parallel lines. 4. In every triangle, check the sum \(180^\circ\). 5. To prove parallelism, find equal alternate interior or corresponding angles. 6. At the end, check that you did not use a property before proving it.
Example 1. Adjacent and Vertical Angles
This example practices the first step: an angle on a straight line and a vertical angle.
Problem. Lines \(AB\) and \(CD\) intersect at \(O\). It is known that \(\angle AOC=68^\circ\). Find the other three angles.
The vertical angle is equal to the given one, so \(\angle BOD=68^\circ\). The angle \(\angle AOD\) is adjacent to \(\angle AOC\), hence \(\angle AOD=180^\circ-68^\circ=112^\circ\). The vertical angle \(\angle BOC\) is also \(112^\circ\).
Comment. In this type of problem, look for vertical angles first, then adjacent angles.
Example 2. Exterior Angle of a Triangle
Here the exterior angle gives the answer without unnecessary computation.
Problem. In triangle \(ABC\), the exterior angle at \(C\) is \(128^\circ\), and \(\angle A=47^\circ\). Find \(\angle B\) and \(\angle C\).
The exterior angle at \(C\) equals the sum of the two remote interior angles: \(128^\circ=\angle A+\angle B\). Therefore \(\angle B=128^\circ-47^\circ=81^\circ\). The interior angle at \(C\) is adjacent to the exterior angle, so \(\angle C=180^\circ-128^\circ=52^\circ\).
Example 3. Transferring Angles Along Parallel Lines
This example shows how a transversal transfers an angle from one line to another.
Problem. Lines \(a\) and \(b\) are parallel. A transversal intersects them at \(A\) and \(B\). One alternate interior angle is \(73^\circ\). Find the other alternate interior angle and the same-side interior angle with it.
For parallel lines, alternate interior angles are equal, so the second such angle is \(73^\circ\). Same-side interior angles sum to \(180^\circ\), hence the required angle is \(180^\circ-73^\circ=107^\circ\).
Example 4. How to Prove Parallelism
Sometimes parallelism is not given; it must be obtained from equal angles.
Problem. Lines \(AC\) and \(BD\) are cut by line \(AB\). It is known that \(\angle CAB=\angle ABD\). Prove that \(AC\parallel BD\).
The angles \(\angle CAB\) and \(\angle ABD\) are alternate interior angles for lines \(AC\) and \(BD\) with transversal \(AB\). They are equal by the condition. By the parallelism criterion, \(AC\parallel BD\).
Example 5. Bisectors of Same-Side Interior Angles
This is a standard construction: the angles sum to \(180^\circ\), so the sum of their halves is \(90^\circ\).
Problem. Two parallel lines are cut by a third line. Prove that the bisectors of two same-side interior angles are perpendicular.
Let the same-side interior angles be \(\alpha\) and \(\beta\). Then \(\alpha+\beta=180^\circ\). Their bisectors make angles \(\frac{\alpha}{2}\) and \(\frac{\beta}{2}\) with the transversal, and the sum of these two angles is \(90^\circ\). In the triangle formed by the two bisectors and the transversal, the third angle is \(90^\circ\). Hence the bisectors are perpendicular.
Example 6. Altitude and Angle Bisector from One Vertex
This example teaches how to express an unknown angle through two angles of a triangle.
Problem. In triangle \(ABC\), altitude \(AH\) and angle bisector \(AL\) are drawn from vertex \(A\). Prove that the angle between them equals half the difference of angles \(B\) and \(C\).
Let \(\angle B=\beta\), \(\angle C=\gamma\), and assume \(\beta>\gamma\). Then \(\angle A=180^\circ-\beta-\gamma\), so \(\angle BAL=\frac{180^\circ-\beta-\gamma}{2}\). Since \(AH\perp BC\), the angle between \(AB\) and \(AH\) is \(90^\circ-\beta\). Therefore the angle between \(AH\) and \(AL\) equals \(\frac{180^\circ-\beta-\gamma}{2}-(90^\circ-\beta)=\frac{\beta-\gamma}{2}\). If \(\gamma>\beta\), we get \(\frac{\gamma-\beta}{2}\).
Example 7. Exterior Bisector and an Isosceles Triangle
Here parallelism is recognised through equal base angles.
Problem. In triangle \(ABC\), \(AB=AC\). Prove that the bisector of the exterior angle at \(A\) is parallel to the base \(BC\).
Let \(\angle B=\angle C=\beta\). Then \(\angle A=180^\circ-2\beta\), and the exterior angle at \(A\) equals \(2\beta\). Its bisector forms an angle \(\beta\) with side \(AB\). This angle is equal to \(\angle ABC\), so by the parallelism criterion the exterior angle bisector is parallel to \(BC\).
Example 8. A Hidden Angle in Parallel Lines
The final example shows how a constructed parallel line turns the problem into ordinary angle chasing.
Problem. In triangle \(ABC\), point \(D\) lies on side \(BC\). Through \(D\), draw lines \(DE\parallel AB\) and \(DF\parallel AC\), where \(E\in AC\), \(F\in AB\). Prove that if \(AD\) is the angle bisector of angle \(A\), then \(AD\) is also the angle bisector of \(\angle EDF\).
Since \(DE\parallel AB\), the angle between \(DE\) and \(AD\) equals the angle between \(AB\) and \(AD\), that is \(\angle BAD\). Since \(DF\parallel AC\), the angle between \(AD\) and \(DF\) equals \(\angle DAC\). By the condition, \(AD\) bisects angle \(A\), so \(\angle BAD=\angle DAC\). Therefore \(AD\) divides \(\angle EDF\) into two equal parts.
Chapter
Triangles I: Congruence
Key Idea
Triangle congruence lets us transfer information: if two triangles are congruent, then their corresponding sides, angles, medians, altitudes, and angle bisectors are equal. In olympiad problems, congruent triangles are often hidden: they must be selected in the diagram or created by an auxiliary line.
The main skill of this module is not only knowing the congruence criteria, but choosing the right pair of triangles and matching corresponding elements accurately.
Basic Facts
SAS: if two sides and the included angle of one triangle are respectively equal to two sides and the included angle of another triangle, then the triangles are congruent. ASA: if a side and the two adjacent angles of one triangle are respectively equal to a side and the two adjacent angles of another triangle, then the triangles are congruent. SSS: if the three sides of one triangle are respectively equal to the three sides of another triangle, then the triangles are congruent.
In an isosceles triangle, the base angles are equal. The median drawn to the base of an isosceles triangle is also an angle bisector and an altitude. Conversely, if two angles of a triangle are equal, then the opposite sides are equal.
When to Use This Method
Look for congruent triangles when you need to prove equality of segments or angles, perpendicularity, midpoint properties, symmetry, or properties of a median or angle bisector. The method is especially common when there is a common side, vertical angles, a midpoint, an angle bisector, an altitude, or equal segments on different rays.
If the needed triangles are not present, try joining two points, extending a side, marking off an equal segment, or drawing a line through a midpoint.
How to Recognise the Method
The problem often asks to prove \(AB=CD\), \(\angle A=\angle D\), that a point is a midpoint, or that a line is an angle bisector or an altitude. This is a signal: find two triangles in which the desired elements become corresponding elements.
A useful habit is to mark equal elements on the diagram first, then search for triangles that already have three elements for one of the congruence criteria.
Typical Mistakes
Do not write “the triangles are congruent” without naming the criterion and the corresponding elements. Do not use SAS if the equal angle is not included between the two equal sides. Do not mix up the order of vertices: \(\triangle ABC=\triangle DEF\) means that \(A\) corresponds to \(D\), \(B\) to \(E\), and \(C\) to \(F\).
Another mistake is proving an element equal while already using that equality as known. The reasons for triangle congruence must come first; the conclusions come after.
Mini-Checklist
1. What must be proved: equality of sides, equality of angles, a midpoint, or perpendicularity? 2. In which two triangles will this become a pair of corresponding elements? 3. Is there a common side or vertical angles? 4. Which criterion fits: SAS, ASA, or SSS? 5. Is the order of corresponding vertices correct? 6. If the triangles are missing, what auxiliary line should be drawn?
Example 1. SAS Criterion
The first example shows how to prove triangle congruence using two sides and the included angle.
Problem. In triangles \(ABC\) and \(A_1B_1C_1\), it is known that \(AB=A_1B_1\), \(AC=A_1C_1\), and \(\angle BAC=\angle B_1A_1C_1\). Prove that \(BC=B_1C_1\).
Triangles \(ABC\) and \(A_1B_1C_1\) are congruent by SAS: two sides and the included angle are respectively equal. Therefore the corresponding third sides are equal: \(BC=B_1C_1\).
Example 2. ASA Criterion
Here it is important to see that the equal side lies between the two equal angles.
Problem. In triangles \(ABC\) and \(DEF\), suppose \(AB=DE\), \(\angle A=\angle D\), and \(\angle B=\angle E\). Prove that \(AC=DF\) and \(BC=EF\).
Side \(AB\) is adjacent to angles \(A\) and \(B\), while side \(DE\) is adjacent to angles \(D\) and \(E\). By ASA, triangles \(ABC\) and \(DEF\) are congruent. Hence the corresponding sides are equal: \(AC=DF\), \(BC=EF\).
Example 3. An Isosceles Triangle
This example connects equal sides with equal angles.
Problem. In triangle \(ABC\), \(AB=AC\), and \(\angle A=44^\circ\). Find angles \(B\) and \(C\).
Since \(AB=AC\), the triangle is isosceles, and \(\angle B=\angle C\). Then \(\angle B+\angle C=180^\circ-44^\circ=136^\circ\), so \(\angle B=\angle C=68^\circ\).
Example 4. Median to the Base
The median in an isosceles triangle gives two properties at once: an angle bisector and an altitude.
Problem. In triangle \(ABC\), \(AB=AC\). Point \(M\) is the midpoint of \(BC\). Prove that \(AM\perp BC\) and \(\angle BAM=\angle MAC\).
Consider triangles \(ABM\) and \(ACM\). We have \(AB=AC\), \(BM=CM\), and \(AM\) is common. By SSS, the triangles are congruent. Hence \(\angle BAM=\angle MAC\), so \(AM\) is an angle bisector. Also, \(\angle AMB=\angle AMC\). These angles are adjacent, so each is \(90^\circ\). Therefore \(AM\perp BC\).
Example 5. Angle Bisector and Altitude
A converse argument: if one line bisects an angle and is also perpendicular to the opposite side, the triangle is isosceles.
Problem. In triangle \(ABC\), the angle bisector \(AD\) of angle \(A\) is perpendicular to \(BC\). Prove that \(AB=AC\).
Triangles \(ABD\) and \(ACD\) are right triangles because \(AD\perp BC\). They have common side \(AD\), and \(\angle BAD=\angle DAC\) because \(AD\) is an angle bisector. By ASA, the triangles are congruent. Therefore the corresponding hypotenuses are equal: \(AB=AC\).
Comment. Here triangle congruence proves isoscelesness, not the other way around.
Example 6. Extending a Median
An auxiliary point often creates a pair of congruent triangles.
Problem. In triangle \(ABC\), point \(M\) is the midpoint of \(BC\). On ray \(AM\) beyond \(M\), choose point \(P\) such that \(MP=AM\). Prove that \(BP=AC\) and \(CP=AB\).
Since \(M\) is the midpoint of \(BC\), \(BM=CM\). Also \(AM=MP\). Angles \(AMB\) and \(PMC\) are vertical, so they are equal. Thus \(\triangle AMB=\triangle PMC\) by SAS, and \(AB=CP\). Similarly, \(\triangle AMC=\triangle PMB\), so \(AC=BP\).
Example 7. Hidden Congruent Triangles
Sometimes congruent triangles appear after subtracting equal segments.
Problem. In isosceles triangle \(ABC\), \(AB=AC\). Points \(D\) and \(E\) lie on sides \(AB\) and \(AC\), respectively, and \(BD=CE\). Prove that \(DE\parallel BC\).
From \(AB=AC\) and \(BD=CE\), we get \(AD=AE\). Hence triangle \(ADE\) is isosceles, so \(\angle ADE=\angle DEA\). Triangle \(ABC\) is also isosceles, so \(\angle ABC=\angle BCA\). Since the two triangles share the angle at \(A\), their base angles are equal. Therefore \(\angle ADE=\angle ABC\), which implies \(DE\parallel BC\).
Example 8. A Kite and a Diagonal
The final example shows how one diagonal can become an axis of symmetry.
Problem. In quadrilateral \(ABCD\), \(AB=AD\) and \(CB=CD\). The diagonals meet at \(O\). Prove that \(AC\perp BD\) and \(BO=DO\).
First consider triangles \(ABC\) and \(ADC\). They have \(AB=AD\), \(CB=CD\), and common side \(AC\), so they are congruent by SSS. Therefore \(\angle BAC=\angle CAD\), meaning that \(AC\) bisects angle \(BAD\). Now in triangles \(ABO\) and \(ADO\), we have \(AB=AD\), common side \(AO\), and \(\angle BAO=\angle OAD\). By SAS, these triangles are congruent, so \(BO=DO\) and \(\angle AOB=\angle AOD\). These angles are adjacent, so each is \(90^\circ\). Hence \(AC\perp BD\).
Chapter
Triangles II: Similarity
Key Idea
Similarity lets us compare figures with the same shape but different size. If two triangles are similar, then their corresponding angles are equal and their corresponding sides are proportional.
In olympiad geometry, similarity often appears when there are parallel lines, common angles, right angles, altitudes, or a small triangle inside a larger one.
Basic Facts
AA: if two angles of one triangle are equal to two angles of another triangle, then the triangles are similar. SAS: if two sides of one triangle are proportional to two sides of another and the included angles are equal, then the triangles are similar. SSS: if the three sides of one triangle are proportional to the three sides of another, then the triangles are similar.
If \(\triangle ABC\sim\triangle A_1B_1C_1\), then \(\frac{AB}{A_1B_1}=\frac{BC}{B_1C_1}=\frac{CA}{C_1A_1}\). This common ratio is called the similarity ratio. The areas of similar triangles are in the square of the similarity ratio.
The midline of a triangle joins the midpoints of two sides. It is parallel to the third side and equals half of it.
When to Use This Method
Look for similarity when the problem contains a parallel line inside a triangle, an altitude in a right triangle, a ratio of segments, scaling, shadows, midlines, or asks for a length through a proportion.
Similarity is especially useful when equality of sides is not enough, but equal angles and proportions are available.
How to Recognise the Method
Common signs of similarity are: two triangles share an angle; one side is drawn parallel to another; right triangles have another common acute angle; proportional segments are marked on the sides of one angle.
A good habit is to write the vertex correspondence first, for example \(\triangle ADE\sim\triangle ABC\), and only then write proportions.
Typical Mistakes
Do not write proportions without the correct vertex correspondence. Do not mix sides from different positions: if \(\triangle ADE\sim\triangle ABC\), then \(AD\) corresponds to \(AB\), \(AE\) to \(AC\), and \(DE\) to \(BC\).
In area problems, students often forget to square the similarity ratio. If lengths are in the ratio \(2:3\), then areas are in the ratio \(4:9\).
Mini-Checklist
1. Which two triangles are being compared? 2. How do the vertices correspond? 3. Are there two equal angles for AA? 4. If sides are used, are they written in the correct order? 5. Are we looking for a length, a ratio, or an area? 6. For areas, remember to square the similarity ratio.
Example 1. Similarity by Two Angles
Basic technique: two equal angles already give similarity.
Problem. In triangles \(ABC\) and \(DEF\), \(\angle A=\angle D\), \(\angle B=\angle E\), \(AB=6\), \(DE=9\), and \(AC=8\). Find \(DF\).
By AA, the triangles are similar: \(\triangle ABC\sim\triangle DEF\). Side \(AB\) corresponds to \(DE\), and \(AC\) corresponds to \(DF\). The scale factor from the first triangle to the second is \(\frac{DE}{AB}=\frac{9}{6}=\frac{3}{2}\). Hence \(DF=8\cdot\frac{3}{2}=12\).
Example 2. A Parallel Line in a Triangle
A parallel side creates a small triangle similar to the large one.
Problem. In triangle \(ABC\), points \(D\in AB\), \(E\in AC\), and \(DE\parallel BC\). It is known that \(AD:DB=2:3\), \(BC=20\). Find \(DE\).
Since \(DE\parallel BC\), we have \(\triangle ADE\sim\triangle ABC\). The ratio \(AD:AB=2:(2+3)=2:5\). Therefore \(\frac{DE}{BC}=\frac{2}{5}\), so \(DE=20\cdot\frac{2}{5}=8\).
Example 3. The Midline
The midline is the first important consequence of similarity and parallel lines.
Problem. In triangle \(ABC\), points \(M\) and \(N\) are the midpoints of sides \(AB\) and \(AC\). Prove that \(MN\parallel BC\) and find \(MN\) if \(BC=14\).
Since \(AM=MB\) and \(AN=NC\), we have \(\frac{AM}{AB}=\frac{AN}{AC}=\frac{1}{2}\). By SAS similarity, triangles \(AMN\) and \(ABC\) are similar. Therefore corresponding sides \(MN\) and \(BC\) are parallel, and \(MN=\frac{1}{2}BC=7\).
Example 4. Areas of Similar Triangles
If lengths increase by a factor of \(k\), area increases by a factor of \(k^2\).
Problem. Triangles \(ABC\) and \(DEF\) are similar, and \(AB:DE=2:5\). The area of \(ABC\) is \(24\). Find the area of \(DEF\).
The ratio of corresponding sides from \(ABC\) to \(DEF\) is \(2:5\), so the areas are in the ratio \(4:25\). Therefore \(S_{DEF}=24\cdot\frac{25}{4}=150\).
Example 5. Altitude in a Right Triangle
The altitude to the hypotenuse creates three similar right triangles.
Problem. In right triangle \(ABC\), angle \(C\) is right, and \(CH\) is the altitude to hypotenuse \(AB\). It is known that \(AH=4\), \(HB=9\). Find \(CH\).
Triangles \(ACH\) and \(CBH\) are similar: both are right triangles, and their acute angles complement each other as angles of the original triangle. From similarity, \(\frac{CH}{AH}=\frac{HB}{CH}\), so \(CH^2=AH\cdot HB=4\cdot9=36\). Hence \(CH=6\).
Example 6. Measuring Height by Shadow
A practical model of similarity: sunlight rays are treated as parallel.
Problem. A vertical stick of height \(1.6\) m casts a shadow of length \(2\) m. At the same moment, a tree casts a shadow of length \(11\) m. Find the height of the tree.
The triangles formed by the stick and its shadow and by the tree and its shadow are similar by two angles: both are right triangles and have the same angle of sunlight. Let the tree height be \(h\). Then \(\frac{h}{11}=\frac{1.6}{2}\), so \(h=8.8\) m.
Example 7. Similarity by Two Sides and an Angle
This example shows that two equal angles are not always needed if side proportions are available.
Problem. In triangles \(ABC\) and \(DEF\), suppose \(\angle A=\angle D\), \(AB:DE=AC:DF=2:3\). Prove that the triangles are similar.
The two sides containing the equal angles are proportional: \(\frac{AB}{DE}=\frac{AC}{DF}\). The included angles are equal. Therefore \(\triangle ABC\sim\triangle DEF\) by SAS similarity.
Example 8. Medians Meet in the Ratio \(2:1\)
Olympiad preparation: the midline helps prove an important property of medians.
Problem. In triangle \(ABC\), medians \(BM\) and \(CN\) meet at \(G\). Prove that \(BG:GM=CG:GN=2:1\).
Let \(P\) and \(Q\) be the midpoints of \(BG\) and \(CG\). Then \(PQ\) is a midline of triangle \(BCG\), so \(PQ\parallel BC\) and \(PQ=\frac{1}{2}BC\). Also, \(MN\) is a midline of triangle \(ABC\), hence \(MN\parallel BC\) and \(MN=\frac{1}{2}BC\). Thus \(PQ\parallel MN\) and \(PQ=MN\), so quadrilateral \(MNPQ\) is a parallelogram. Its diagonals bisect each other, so \(G\) is the midpoint of \(MP\) and \(NQ\). Since \(P\) is the midpoint of \(BG\), we get \(BG=2GP=2GM\). Similarly, \(CG=2GN\).
Chapter
Quadrilaterals
Key Idea
A quadrilateral is rarely solved as one whole figure. Usually we split it into two triangles by a diagonal, use parallel lines, or examine the intersection of the diagonals.
In olympiad problems it is important not only to know the properties of a parallelogram, rhombus, or trapezoid, but also to prove that a given quadrilateral has the required type.
Basic Facts
A parallelogram is a quadrilateral whose two pairs of opposite sides are parallel. In a parallelogram, opposite sides are equal, opposite angles are equal, and the diagonals bisect each other.
Useful parallelogram criteria: both pairs of opposite sides are parallel; both pairs of opposite sides are equal; one pair of opposite sides is both equal and parallel; the diagonals bisect each other.
A rectangle is a parallelogram with a right angle. A rhombus is a parallelogram with equal adjacent sides. A square is both a rectangle and a rhombus.
In a trapezoid, one pair of opposite sides is parallel. If the bases of a trapezoid are \(a\) and \(b\), then its midline equals \(\frac{a+b}{2}\) and is parallel to the bases.
Circle preview: if in a convex quadrilateral the sum of opposite angles is \(180^\circ\), then its vertices lie on one circle. Conversely, in a cyclic quadrilateral, opposite angles are supplementary.
When to Use This Method
Use quadrilateral properties when a problem contains two pairs of parallel lines, midpoints of sides, diagonals, a trapezoid, equal bases, or asks to prove parallelism.
If you need to prove that a quadrilateral is a parallelogram, it is often easier to check the diagonals or one pair of equal and parallel opposite sides than to prove two pairs of parallel sides directly.
How to Recognise the Method
Signs of a parallelogram: an intersection point of diagonals, midpoints, symmetric segments, two parallel lines, and equal segments on them. Signs of a trapezoid: one pair of parallel sides, midpoints of the legs, diagonals, and a line parallel to the bases.
Signs of a cyclic quadrilateral: equal angles standing on the same segment, or a sum of opposite angles equal to \(180^\circ\). In this module this is only a preview tool, but it already helps in proofs.
Typical Mistakes
Do not treat a quadrilateral as a parallelogram just because the sides look parallel in a drawing. Do not use rectangle properties before proving a right angle or equal diagonals in a parallelogram.
In a trapezoid, do not confuse the bases with the legs. The midline joins the midpoints of the legs, not arbitrary points on the sides.
For a cyclic quadrilateral, one pair of equal angles is not always enough by itself: you must understand which segment those angles stand on, or use the sum of opposite angles.
Mini-Checklist
1. Which sides are parallel? 2. Are there midpoints or diagonals? 3. Can the parallelogram be proved through its diagonals? 4. Is the parallelogram actually a rectangle or a rhombus? 5. In a trapezoid, have the midpoints of the legs been found? 6. For a circle, has the sum of opposite angles \(180^\circ\) been checked?
Example 1. A Parallelogram Through Diagonals
Basic idea: if the diagonals bisect each other, this is already a strong parallelogram criterion.
Problem. In quadrilateral \(ABCD\), diagonals \(AC\) and \(BD\) meet at \(O\). It is known that \(AO=OC\) and \(BO=OD\). Prove that \(ABCD\) is a parallelogram.
Point \(O\) is the midpoint of both diagonals. By the parallelogram criterion, if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. Hence \(ABCD\) is a parallelogram.
Comment. This criterion is often the shortest route to proving a parallelogram.
Example 2. One Pair of Sides Equal and Parallel
A second standard criterion: one pair of opposite sides is enough if it is both equal and parallel.
Problem. In quadrilateral \(ABCD\), suppose \(AB\parallel CD\) and \(AB=CD\). Prove that \(ABCD\) is a parallelogram.
Draw diagonal \(AC\). Since \(AB\parallel CD\), we have \(\angle BAC=\angle ACD\). Also, \(AC\) is common and \(AB=CD\). Triangles \(BAC\) and \(DCA\) are congruent by two sides and the included angle. Therefore \(\angle BCA=\angle CAD\), so \(BC\parallel AD\). Thus both pairs of opposite sides are parallel, and \(ABCD\) is a parallelogram.
Example 3. When a Parallelogram Becomes a Rectangle
Equal diagonals in a parallelogram force a right angle.
Problem. In parallelogram \(ABCD\), the diagonals are equal: \(AC=BD\). Prove that \(ABCD\) is a rectangle.
Consider triangles \(ABC\) and \(DCB\). In a parallelogram, \(AB=CD\); side \(BC\) is common; and by condition \(AC=BD\). Hence the triangles are congruent by SSS, so \(\angle ABC=\angle DCB\). But angles \(ABC\) and \(DCB\) are same-side interior angles for the parallel lines \(AB\parallel CD\), so their sum is \(180^\circ\). Equal angles with sum \(180^\circ\) are each \(90^\circ\). Therefore the parallelogram is a rectangle.
Example 4. When a Parallelogram Becomes a Rhombus
Perpendicular diagonals in a parallelogram force adjacent sides to be equal.
Problem. In parallelogram \(ABCD\), diagonals \(AC\) and \(BD\) are perpendicular. Prove that \(ABCD\) is a rhombus.
Let the diagonals meet at \(O\). In a parallelogram the diagonals bisect each other, so \(BO=OD\). Triangles \(AOB\) and \(AOD\) are right triangles, have common leg \(AO\), and have equal legs \(BO\) and \(OD\). Thus they are congruent, so \(AB=AD\). In a parallelogram, equality of adjacent sides means that all sides are equal. Hence \(ABCD\) is a rhombus.
Example 5. Midline of a Trapezoid
The midline of a trapezoid is conveniently proved through two triangle midlines.
Problem. In trapezoid \(ABCD\), bases \(AD\parallel BC\). Points \(M\) and \(N\) are the midpoints of legs \(AB\) and \(CD\). Prove that \(MN\parallel AD\parallel BC\), and find \(MN\) if \(AD=16\), \(BC=10\).
Let \(P\) be the midpoint of diagonal \(AC\). In triangle \(ABC\), segment \(MP\) is a midline, so \(MP\parallel BC\) and \(MP=\frac{1}{2}BC\). In triangle \(ACD\), segment \(PN\) is a midline, so \(PN\parallel AD\) and \(PN=\frac{1}{2}AD\). Since \(AD\parallel BC\), points \(M,P,N\) lie on one line. Thus \(MN=MP+PN=\frac{BC+AD}{2}=13\).
Example 6. First Criterion for a Cyclic Quadrilateral
A cyclic quadrilateral is conveniently recognised by the sum of opposite angles.
Problem. In a convex quadrilateral \(ABCD\), \(\angle ABC+\angle ADC=180^\circ\). Prove that points \(A,B,C,D\) lie on one circle.
Draw the circle through points \(A,B,C\). For any point \(D\) on the appropriate arc of this circle, angle \(\angle ADC\) supplements angle \(\angle ABC\) to \(180^\circ\). This is exactly what the condition gives, so point \(D\) lies on the same circle. Therefore \(A,B,C,D\) form a cyclic quadrilateral.
Comment. This is a preview of a circle criterion; in the next module it will become a main tool.
Example 7. Diagonals of a Trapezoid and the Ratio of Bases
In a trapezoid, the intersection of diagonals divides them in the ratio of the bases.
Problem. In trapezoid \(ABCD\), bases \(AD\parallel BC\), \(AD=18\), \(BC=12\). The diagonals meet at \(O\). Find \(AO:OC\) and \(DO:OB\).
Triangles \(AOD\) and \(COB\) are similar: the angles at \(O\) are vertical, and the other corresponding angles are equal because \(AD\parallel BC\). Therefore \(\frac{AO}{OC}=\frac{DO}{OB}=\frac{AD}{BC}=\frac{18}{12}=\frac{3}{2}\). Hence \(AO:OC=DO:OB=3:2\).
Example 8. A Parallelogram From Side Midpoints
A classic olympiad trick: the side midpoints of any quadrilateral form a parallelogram.
Problem. In a convex quadrilateral \(ABCD\), points \(M,N,P,Q\) are the midpoints of sides \(AB,BC,CD,DA\), respectively. Prove that \(MNPQ\) is a parallelogram. If \(AC=BD\), prove that \(MNPQ\) is a rhombus.
In triangle \(ABC\), segment \(MN\) is a midline, so \(MN\parallel AC\) and \(MN=\frac{1}{2}AC\). In triangle \(CDA\), segment \(PQ\) is a midline, so \(PQ\parallel AC\) and \(PQ=\frac{1}{2}AC\). Hence \(MN\parallel PQ\) and \(MN=PQ\), so \(MNPQ\) is a parallelogram. Similarly, \(NP=\frac{1}{2}BD\). If \(AC=BD\), then adjacent sides \(MN\) and \(NP\) are equal, so parallelogram \(MNPQ\) is a rhombus.
Chapter
Circles I: Basic Circle Geometry
Key Idea
A circle turns equal arcs and chords into equal angles. The most important pattern in this module is: if points \(A,B,C,D\) lie on one circle, then angles \(\angle ABC\) and \(\angle ADC\), standing on the same chord \(AC\), are equal or supplementary depending on the positions of the points.
At the first olympiad level, circles are most often used for angle chasing: to find or prove equal angles, to prove that four points lie on one circle, or to use a tangent.
Basic Facts
A central angle is twice an inscribed angle standing on the same arc: if \(O\) is the centre of the circle, then \(\angle AOB=2\angle ACB\) when both angles stand on arc \(AB\).
Inscribed angles standing on the same chord or arc are equal. If \(A,B,C,D\) lie on one circle, we can often write \(\angle ABC=\angle ADC\), because both angles look at chord \(AC\).
An angle standing on a diameter is right. Equal chords cut off equal arcs and give equal inscribed angles. The radius drawn to the point of tangency is perpendicular to the tangent.
The angle between the tangent at \(A\) and chord \(AB\) equals the inscribed angle standing on chord \(AB\) on the other side of the circle.
A quadrilateral is cyclic if and only if the sum of its opposite angles is \(180^\circ\), or when two points see the same segment under equal angles.
When to Use This Method
Look for a circle when a problem contains equal angles, right angles, a tangent, chords, a centre of a circle, a diameter, or the phrase “prove that the points lie on one circle”.
If you need to prove equality of angles, check whether they stand on the same segment. If you need to prove cyclicity, look either for opposite angles summing to \(180^\circ\) or two equal angular views of the same segment.
How to Recognise the Method
Circle signals include angles of the form \(\angle ABC\) and \(\angle ADC\), two altitudes giving right angles, a tangent and a chord, equal chords, a centre, and radii.
A common move is to first prove that four points lie on one circle, and then replace one angle by an equal angle that is easier to connect to the rest of the figure.
Typical Mistakes
Do not use equality of inscribed angles before proving that the points lie on one circle. Do not confuse a central angle with an inscribed angle: the central angle is twice as large, not equal.
In the tangent-chord theorem, make sure the angle is between the tangent and that particular chord. In cyclic quadrilateral problems, track which angles are opposite and which stand on the same chord.
Mini-Checklist
1. Which points already lie on one circle? 2. Which chord does the angle stand on? 3. Is there a diameter or a right angle? 4. Is there a tangent and a radius to the point of tangency? 5. Can cyclicity be proved by a sum of \(180^\circ\)? 6. Can an angle be replaced by an equal angle on the same chord?
Example 1. Central and Inscribed Angle
Basic technique: an inscribed angle is half the central angle standing on the same arc.
Problem. Points \(A,B,C\) lie on a circle with centre \(O\). It is known that \(\angle AOB=124^\circ\). Find \(\angle ACB\), if points \(O\) and \(C\) lie on opposite sides of chord \(AB\).
Angle \(\angle AOB\) is central, and \(\angle ACB\) is inscribed; both stand on arc \(AB\). Therefore \(\angle ACB=\frac{1}{2}\angle AOB=62^\circ\).
Example 2. Angles on the Same Chord
The main pattern of the module: angles looking at the same chord are equal.
Problem. Points \(A,B,C,D\) lie on one circle. Prove that \(\angle ABC=\angle ADC\), if points \(B\) and \(D\) lie on the same side of chord \(AC\).
Both angles \(\angle ABC\) and \(\angle ADC\) are inscribed and stand on the same chord \(AC\). With the given position, they look at the same arc \(AC\). Therefore \(\angle ABC=\angle ADC\).
Example 3. A Diameter Gives a Right Angle
If an angle stands on a diameter, it is right.
Problem. \(AB\) is a diameter of a circle, and point \(C\) lies on the circle. Prove that \(\angle ACB=90^\circ\).
The central angle standing on diameter \(AB\) equals \(180^\circ\). The inscribed angle \(\angle ACB\) standing on the same arc is half as large. Hence \(\angle ACB=90^\circ\).
Example 4. Equal Chords
Equal chords give equal arcs, and therefore equal inscribed angles.
Problem. Points \(A,B,C,D\) lie on a circle, and \(AB=CD\). Prove that \(\angle ADB=\angle CAD\).
Equal chords \(AB\) and \(CD\) cut off equal arcs. Angle \(\angle ADB\) stands on chord \(AB\), and angle \(\angle CAD\) stands on chord \(CD\). Since the corresponding arcs are equal, these inscribed angles are equal.
Example 5. Tangent and Radius
The radius to the point of tangency is always perpendicular to the tangent.
Problem. Line \(t\) is tangent to a circle with centre \(O\) at point \(A\). Prove that \(OA\perp t\).
If a shorter perpendicular from \(O\) to line \(t\) landed at another point, that point would be inside the circle and the line would meet the circle in two points. But \(t\) touches the circle only at \(A\). Therefore the shortest distance from \(O\) to \(t\) is attained at \(A\), so \(OA\perp t\).
Example 6. Angle Between a Tangent and a Chord
A tangent lets us replace an external angle by an internal inscribed angle.
Problem. A tangent is drawn at point \(A\) to the circumcircle of triangle \(ABC\). The angle between the tangent and chord \(AB\) is \(48^\circ\). Find \(\angle ACB\).
By the tangent-chord theorem, the angle between the tangent at \(A\) and chord \(AB\) equals the inscribed angle standing on chord \(AB\). That angle is \(\angle ACB\). Hence \(\angle ACB=48^\circ\).
Example 7. Proving Cyclicity
Four points often lie on one circle because of two right angles.
Problem. In triangle \(ABC\), points \(D\) and \(E\) are the feet of the altitudes from \(B\) and \(C\), respectively. Prove that points \(B,C,D,E\) lie on one circle.
Since \(BD\perp AC\), and \(D\) lies on \(AC\), we have \(\angle BDC=90^\circ\). Since \(CE\perp AB\), and \(E\) lies on \(AB\), we have \(\angle BEC=90^\circ\). Thus points \(D\) and \(E\) see segment \(BC\) under a right angle, so they lie on the circle with diameter \(BC\). Therefore \(B,C,D,E\) lie on one circle.
Example 8. Cyclic Quadrilateral and Sum of Angles
In a cyclic quadrilateral, opposite angles sum to \(180^\circ\).
Problem. In cyclic quadrilateral \(ABCD\), it is known that \(\angle A=73^\circ\). Find \(\angle C\).
Opposite angles of a cyclic quadrilateral sum to \(180^\circ\). Therefore \(\angle C=180^\circ-73^\circ=107^\circ\).
Example 9. Tangents From One Point
Perpendicularity of the radius to the tangent helps obtain congruent right triangles.
Problem. From point \(T\), tangents \(TA\) and \(TB\) are drawn to a circle with centre \(O\). Prove that \(TA=TB\).
Radii \(OA\) and \(OB\) are perpendicular to tangents \(TA\) and \(TB\). Triangles \(OTA\) and \(OTB\) are right triangles, have common hypotenuse \(OT\), and equal legs \(OA=OB\). Hence the triangles are congruent, so \(TA=TB\).
Example 10. A Mixed Circle and Tangent Problem
Olympiad preparation: a tangent often gives an angle that is then used in a cyclic configuration.
Problem. In triangle \(ABC\), the tangent to its circumcircle at \(A\) is parallel to \(BC\). Prove that \(AB=AC\).
The angle between the tangent at \(A\) and chord \(AB\) equals \(\angle ACB\). But the tangent is parallel to \(BC\), so the same angle equals \(\angle ABC\). Therefore \(\angle ABC=\angle ACB\), and triangle \(ABC\) is isosceles: \(AB=AC\).
Chapter
Areas I
Key Idea
Area often replaces long arguments with angles and similarity. If two triangles have a common height, their areas are in the ratio of their bases. If they have a common base, their areas are in the ratio of their heights.
In olympiad problems, area is useful as a “weight”: split a figure into parts, compare the parts, and add or subtract equal areas.
Basic Facts
The area of a triangle is \(S=\frac{1}{2}ah\), where \(a\) is the chosen base and \(h\) is the height to it.
If two triangles have equal bases and equal heights, then their areas are equal. If their heights are equal, their areas are in the ratio of their bases. If their bases are equal, their areas are in the ratio of their heights.
A median divides a triangle into two equal-area triangles, because the two bases on one side are equal and the height to that side is common.
If point \(D\) lies on side \(BC\) of triangle \(ABC\), then \(S_{ABD}:S_{ACD}=BD:DC\). This is one of the main patterns of the module.
In a parallelogram, a diagonal halves the area. In a trapezoid, diagonals and midlines are often studied through areas of triangles whose bases lie on parallel lines.
When to Use This Method
Try areas when a problem contains midpoints, medians, points on one side, parallel lines, ratios of segments, or asks to prove equality of segments without obvious similarity.
Areas are especially useful when several triangles have the same height, when vertices lie on a line parallel to the base, or when a figure is split by intersecting segments.
How to Recognise the Method
Look for a common base or a common height. If two vertices lie on a line parallel to the base, their heights to this base are equal. If a base is divided in the ratio \(m:n\), then the areas of triangles with the same opposite vertex are also in the ratio \(m:n\).
For area chasing, it is convenient to denote several small areas by letters and write the equalities produced by medians, parallel lines, or common heights.
Typical Mistakes
Do not compare areas just from the drawing. You must explicitly state that the bases are equal, the heights are equal, or the heights are in a known ratio.
Do not confuse area ratios with side ratios in similar triangles: if similarity is used, areas are in the square of the similarity ratio. If a common height is used, areas are in the ratio of the bases.
In area chasing, add and subtract areas of the same regions. Do not subtract equalities unless it is clear which regions each area contains.
Mini-Checklist
1. Which base should be chosen? 2. Is there a common height? 3. Are there equal bases or midpoints? 4. Can an area ratio be replaced by a segment ratio? 5. Which small regions should be named? 6. Should equal areas be added or subtracted?
Example 1. Area of a Triangle
Basic technique: choose a base and the height to it.
Problem. A triangle has base \(13\) and height to this base \(8\). Find its area.
By the triangle area formula, \(S=\frac{1}{2}ah\). Therefore \(S=\frac{1}{2}\cdot13\cdot8=52\).
Example 2. One Height, Different Bases
If the opposite vertex is common and the bases lie on one line, the height is the same.
Problem. In triangle \(ABC\), point \(D\) lies on side \(BC\), and \(BD:DC=2:5\). Find \(S_{ABD}:S_{ACD}\).
Triangles \(ABD\) and \(ACD\) have a common height from point \(A\) to line \(BC\). Therefore their areas are in the ratio of their bases: \(S_{ABD}:S_{ACD}=BD:DC=2:5\).
Example 3. A Median Halves Area
A median is one of the most common sources of equal areas.
Problem. In triangle \(ABC\), point \(M\) is the midpoint of \(BC\). Prove that \(S_{ABM}=S_{ACM}\).
Triangles \(ABM\) and \(ACM\) have equal bases \(BM\) and \(CM\), and the height from point \(A\) to line \(BC\) is common. Hence their areas are equal.
Example 4. Equal Heights Between Parallel Lines
Parallel lines often give equal heights.
Problem. Points \(A\) and \(B\) lie on one line, and points \(C\) and \(D\) lie on a line parallel to \(AB\). Prove that \(S_{ABC}=S_{ABD}\).
Triangles \(ABC\) and \(ABD\) have common base \(AB\). The heights from points \(C\) and \(D\) to line \(AB\) are equal because \(C\) and \(D\) lie on a line parallel to \(AB\). Therefore the areas are equal.
Example 5. A Point on a Median
Every point on a median gives equal areas with the two halves of the base.
Problem. In triangle \(ABC\), median \(AM\) is drawn to side \(BC\). Point \(P\) lies on \(AM\). Prove that \(S_{PAB}=S_{PAC}\).
Since \(P\) lies on median \(AM\), line \(AP\) passes through the midpoint \(M\) of side \(BC\). Triangles \(PAB\) and \(PAC\) have common base \(AP\). Points \(B\) and \(C\) are equally distant from line \(AP\), because \(M\) is the midpoint of \(BC\). Hence the areas are equal.
Example 6. Diagonal of a Parallelogram
A parallelogram is conveniently cut by a diagonal into two equal-area triangles.
Problem. Prove that diagonal \(AC\) of parallelogram \(ABCD\) halves its area.
Triangles \(ABC\) and \(ACD\) have common base \(AC\). Since \(AB\parallel CD\), the heights from points \(B\) and \(D\) to line \(AC\) are equal. Therefore \(S_{ABC}=S_{ACD}\), and the diagonal halves the area of the parallelogram.
Example 7. Area Division Inside a Triangle
If a point moves along a segment toward the base, its height to the base changes linearly.
Problem. Triangle \(ABC\) has area \(90\). Point \(D\) lies on \(BC\), and point \(E\) lies on \(AD\), with \(AE:ED=1:2\). Find \(S_{BCE}\).
Triangles \(BCE\) and \(BCA\) have common base \(BC\). The height of point \(E\) to \(BC\) is \(\frac{2}{3}\) of the height of point \(A\), because \(AE:ED=1:2\). Therefore \(S_{BCE}=\frac{2}{3}S_{ABC}=60\).
Example 8. Six Equal Areas
The medians of a triangle divide it into six equal-area small triangles.
Problem. In triangle \(ABC\), the medians meet at point \(G\). Prove that the six small triangles around \(G\) have equal areas.
The intersection point of the medians divides each median in the ratio \(2:1\). Consider two small regions on opposite sides of one median: they have equal bases on a side of the triangle and a common height from \(G\). This gives pairs of equal areas. Also, each median halves the whole triangle. Comparing the halves, we get that all six small areas are equal.
Example 9. Areas in a Trapezoid
In a trapezoid, it is useful to compare triangles with bases on parallel lines.
Problem. In trapezoid \(ABCD\), bases \(AD\parallel BC\), and the diagonals meet at point \(O\). Prove that \(S_{AOB}=S_{COD}\).
Triangles \(ABD\) and \(ACD\) have common base \(AD\), and the heights from points \(B\) and \(C\) to \(AD\) are equal because \(BC\parallel AD\). Hence \(S_{ABD}=S_{ACD}\). Subtracting the common area \(S_{AOD}\), we obtain \(S_{AOB}=S_{COD}\).
Example 10. Area Chasing Through an Interior Point
If three small areas are equal, the point lies on the medians.
Problem. Point \(P\) lies inside triangle \(ABC\), and \(S_{PAB}=S_{PAC}=S_{PBC}\). Prove that \(P\) is the intersection point of the medians.
From \(S_{PAB}=S_{PAC}\), points \(B\) and \(C\) have equal distances to line \(AP\). Hence line \(AP\) passes through the midpoint of \(BC\), so it is a median. Similarly, from \(S_{PAB}=S_{PBC}\), line \(BP\) is a median. The intersection of two medians is the centroid, so \(P\) is the intersection point of the medians.
Chapter
Basic Constructions and Auxiliary Lines
Key Idea
An auxiliary construction should not decorate the diagram. It should create a familiar situation: congruent triangles, parallel lines, a midpoint, a parallelogram, similarity, or a circle.
This module works as a toolbox: the student learns to choose a move, not to memorise one theorem. After each construction, immediately ask: what new equal angles, equal segments, or congruent triangles appeared?
Basic Facts
Extending a side is useful when we need an exterior angle, need to construct an equal segment, or need to create a triangle congruent to one already present.
A parallel line is useful when we need equal angles, a midline, similar triangles, or a parallelogram. If a line through the midpoint of one side of a triangle is drawn parallel to another side, a new midpoint often appears.
A circle is useful when there is a right angle, equal angles, or a need to prove that four points lie on one circle. A common construction is the circle with a given segment as diameter.
Reflecting a point about a midpoint or extending a median by an equal segment often creates a parallelogram and hidden congruent triangles.
When to Use This Method
Add a point if the problem lacks the second side of a congruent triangle, if a parallelogram is not yet visible, or if a segment needs to be “moved” to another place.
Draw a parallel line if the diagram contains a midpoint, a ratio on a side, a trapezoid, or a need for similarity. Draw a circle if there are two right angles, equal angles on one segment, or a tangent.
How to Recognise the Method
A midpoint often asks you to extend a segment by the same length. A ratio on a side often asks for a parallel line. Two right angles often ask for a circle with a diameter. Equal segments often ask for a circle or an isosceles triangle.
If after a construction you cannot name a new fact, the construction was probably chosen at random.
Typical Mistakes
Do not draw many lines without a purpose. Do not use a property of a constructed figure before proving it: for example, do not call a quadrilateral a parallelogram just because it looks like one.
When extending a side, state the order of the points. When drawing a circle, explain why the needed points lie on it. When drawing a parallel line, write down the equal angles it creates.
Mini-Checklist
1. What object do I want to create: a congruent triangle, parallelogram, similarity, or circle? 2. Where is there a midpoint? 3. Can a segment be extended by an equal length? 4. Which parallel line gives the needed angles? 5. Is there a diameter or two right angles? 6. What exactly became true after the construction?
Example 1. Extend a Median
Tool: a midpoint often asks us to extend a segment by the same length.
Problem. In triangle \(ABC\), point \(M\) is the midpoint of \(BC\). Extend \(AM\) beyond \(M\) to point \(D\) so that \(MD=AM\). Prove that \(AB\parallel CD\) and \(AC\parallel BD\).
Point \(M\) is the midpoint of both \(BC\) and \(AD\). Therefore the diagonals of quadrilateral \(ABDC\) bisect each other. Hence \(ABDC\) is a parallelogram. Thus \(AB\parallel CD\) and \(AC\parallel BD\).
Example 2. Draw a Parallel Through a Midpoint
Tool: a parallel line turns the midpoint of one side into the midpoint of another.
Problem. In triangle \(ABC\), point \(D\) is the midpoint of \(AB\). Through \(D\), a line parallel to \(BC\) meets \(AC\) at \(E\). Prove that \(E\) is the midpoint of \(AC\).
Since \(DE\parallel BC\), triangles \(ADE\) and \(ABC\) are similar. From \(AD=DB\), we get \(AD:AB=1:2\). Therefore \(AE:AC=1:2\), so \(AE=EC\). Thus \(E\) is the midpoint of \(AC\).
Example 3. Complete a Parallelogram
Tool: if equal sides are missing, it is often useful to complete a parallelogram.
Problem. Through points \(B\) and \(C\) of triangle \(ABC\), draw lines parallel to \(AC\) and \(AB\), respectively; they meet at point \(D\). Prove that \(AB=CD\) and \(AC=BD\).
By construction, \(BD\parallel AC\) and \(CD\parallel AB\). Hence \(ABDC\) is a parallelogram. Opposite sides of a parallelogram are equal, so \(AB=CD\) and \(AC=BD\).
Example 4. Circle With a Diameter
Tool: two right angles often indicate a circle with a diameter.
Problem. In quadrilateral \(ABCD\), \(\angle ACB=90^\circ\) and \(\angle ADB=90^\circ\). Prove that points \(A,B,C,D\) lie on one circle.
Draw the circle with diameter \(AB\). Every point from which segment \(AB\) is seen under a right angle lies on this circle. Therefore points \(C\) and \(D\) lie on the circle with diameter \(AB\), so \(A,B,C,D\) lie on one circle.
Example 5. Construct an Equal Segment
Tool: an equal segment creates an isosceles triangle.
Problem. In triangle \(ABC\), extend ray \(BA\) beyond point \(B\) to point \(D\) so that \(BD=BC\). If \(\angle ABC=44^\circ\), find \(\angle BCD\).
Since \(D\) lies on ray \(BA\), angle \(\angle DBC\) equals \(\angle ABC=44^\circ\). In triangle \(BCD\), sides \(BD\) and \(BC\) are equal, so the base angles are equal. Hence \(\angle BCD=\frac{180^\circ-44^\circ}{2}=68^\circ\).
Example 6. Reflection About a Midpoint
Tool: reflecting a point about a midpoint often gives a parallelogram.
Problem. In triangle \(ABC\), point \(M\) is the midpoint of \(BC\). Point \(D\) is the reflection of \(A\) about \(M\). Prove that \(ABDC\) is a parallelogram.
From the reflection, \(AM=MD\), and by condition \(BM=MC\). Thus diagonals \(AD\) and \(BC\) of quadrilateral \(ABDC\) are bisected by point \(M\). Therefore \(ABDC\) is a parallelogram.
Example 7. Circle Through Three Points
Tool: if two points see one segment under equal angles, a circle appears.
Problem. Points \(C\) and \(D\) lie on the same side of line \(AB\), and \(\angle ACB=\angle ADB\). Prove that points \(A,B,C,D\) lie on one circle.
Draw the circle through \(A,B,C\). All points from which segment \(AB\) is seen under angle \(\angle ACB\) on the same side of \(AB\) lie on one arc of this circle. Since \(\angle ADB=\angle ACB\), point \(D\) lies on the same circle.
Example 8. A Diagonal as an Auxiliary Line
Tool: in a quadrilateral, trying a diagonal is almost always worthwhile.
Problem. In convex quadrilateral \(ABCD\), points \(M,N,P,Q\) are the midpoints of sides \(AB,BC,CD,DA\). Prove that \(MNPQ\) is a parallelogram.
Draw diagonal \(AC\). In triangle \(ABC\), segment \(MN\) is a midline, so \(MN\parallel AC\). In triangle \(CDA\), segment \(PQ\) is a midline, so \(PQ\parallel AC\). Hence \(MN\parallel PQ\). Similarly, drawing diagonal \(BD\), we get \(NP\parallel MQ\). Therefore \(MNPQ\) is a parallelogram.
Chapter
Mixed Problems I
Key Idea
In a mixed problem, the method is not written in the statement. First read the configuration: where are the parallel lines, equal segments, midpoints, right angles, circles, ratios, and areas?
A good olympiad habit is not to start with computation. First find the structure: an angle on a chord, a hidden congruent triangle, a small triangle inside a large one, a median, a common height, or a useful auxiliary construction.
Basic Facts
Angle chasing works when there are parallel lines, an exterior angle, the angle sum of a triangle, or a cyclic quadrilateral.
Triangle congruence is found by SSS, SAS, ASA, especially after extending a side, reflecting a point, or completing a parallelogram.
Similarity appears from parallel lines, common angles, altitudes in a right triangle, and proportions. Areas are useful with common heights, medians, points on one side, and decompositions of a figure.
Circles help replace angles: if points lie on one circle, angles standing on one chord are equal; if two angles are right, there is often a circle with a diameter.
When to Use This Method
If you need to find an angle, first look for parallel lines and a circle. If you need to prove equality of segments, look for congruent triangles, equal chords, or a parallelogram. If you need to find a ratio, look for similarity or areas.
If the diagram does not give the needed pair of triangles, try drawing a diagonal, extending a median, drawing a parallel line, or constructing a circle with a diameter.
How to Recognise the Method
A midpoint suggests a median, a midline, areas, or extending by an equal segment. Parallelism suggests angles and similarity. Right angles suggest a circle with a diameter. A ratio on a side suggests similarity or areas.
If the problem contains a quadrilateral, try a diagonal or a circle. If there is an intersection of diagonals, try ratios and areas.
Typical Mistakes
Do not choose a method just because it was studied most recently. In a mixed block, a problem may look like angles but be solved by areas, or look like areas but be solved by similarity.
Do not use a property before proving the figure has it: cyclicity, parallelogram structure, isosceles triangles, and similarity must be justified. Do not overload the diagram with lines: every construction should create a concrete new fact.
Mini-Checklist
1. What is required: an angle, segment, ratio, area, or proof? 2. Are there parallel lines? 3. Are there midpoints or medians? 4. Is there a circle or two right angles? 5. Can triangles be compared? 6. Would areas be simpler? 7. Which one auxiliary construction creates a familiar situation?
Example 1. First Recognise the Angles
This example shows that a parallel line often turns the problem into angle chasing.
Problem. In triangle \(ABC\), point \(D\) lies on \(AC\), and \(DE\parallel BC\), where \(E\) lies on \(AB\). If \(\angle A=48^\circ\), \(\angle B=67^\circ\), find \(\angle ADE\).
Since \(DE\parallel BC\), angle \(\angle ADE\) equals \(\angle ACB\). In triangle \(ABC\), \(\angle C=180^\circ-48^\circ-67^\circ=65^\circ\). Therefore \(\angle ADE=65^\circ\).
Example 2. A Midpoint Asks for an Extension
If there is a midpoint and congruent triangles are missing, extending by an equal segment helps.
Problem. In triangle \(ABC\), point \(M\) is the midpoint of \(BC\). Extend \(AM\) beyond \(M\) to \(D\) so that \(MD=AM\). Prove that \(AB\parallel CD\).
We have \(AM=MD\), \(BM=MC\), and \(\angle AMB=\angle DMC\) as vertical angles. Therefore \(\triangle ABM\cong\triangle DCM\). Hence \(\angle ABM=\angle DCM\), so \(AB\parallel CD\).
Example 3. Ratio Through Similarity
If a parallel line is drawn inside a triangle, similarity is almost always worth checking.
Problem. In triangle \(ABC\), points \(D\) and \(E\) lie on \(AB\) and \(AC\), and \(DE\parallel BC\). If \(AD:DB=3:2\) and \(BC=20\), find \(DE\).
From \(DE\parallel BC\), \(\triangle ADE\sim\triangle ABC\). We have \(AD:AB=3:5\), hence \(DE:BC=3:5\). Therefore \(DE=12\).
Example 4. A Circle Replaces an Angle
If four points are cyclic, one angle can be replaced by another.
Problem. Points \(A,B,C,D\) lie on one circle, and \(B\) and \(D\) are on the same side of chord \(AC\). If \(\angle ABC=42^\circ\), find \(\angle ADC\).
Angles \(\angle ABC\) and \(\angle ADC\) stand on the same chord \(AC\). With the given position, they are equal. Therefore \(\angle ADC=42^\circ\).
Example 5. Areas Instead of Angles
Sometimes a segment ratio is easier to get from areas.
Problem. In triangle \(ABC\), point \(D\) lies on \(BC\). It is known that \(S_{ABD}=18\), \(S_{ACD}=30\). Find \(BD:DC\).
Triangles \(ABD\) and \(ACD\) have a common height from \(A\) to \(BC\). Therefore \(BD:DC=S_{ABD}:S_{ACD}=18:30=3:5\).
Example 6. A Diagonal of a Quadrilateral
In a quadrilateral, a diagonal often creates midlines or congruent triangles.
Problem. In quadrilateral \(ABCD\), points \(M,N,P,Q\) are the side midpoints. Prove that \(MNPQ\) is a parallelogram.
Draw diagonal \(AC\). Then \(MN\parallel AC\) and \(PQ\parallel AC\), because they are midlines in triangles \(ABC\) and \(CDA\). Hence \(MN\parallel PQ\). Similarly, using diagonal \(BD\), we get \(NP\parallel MQ\). Therefore \(MNPQ\) is a parallelogram.
Example 7. Two Right Angles Give a Circle
If two points see one segment under a right angle, a circle with a diameter appears.
Problem. In triangle \(ABC\), the feet of the altitudes from \(B\) and \(C\) are \(D\) and \(E\). Prove that \(B,C,D,E\) lie on one circle.
We have \(\angle BDC=90^\circ\) and \(\angle BEC=90^\circ\). Therefore points \(D\) and \(E\) lie on the circle with diameter \(BC\). Hence \(B,C,D,E\) are cyclic.
Example 8. A Mixed Move in a Trapezoid
Here both similarity and understanding of trapezoid diagonals are needed.
Problem. In trapezoid \(ABCD\), bases \(AD\parallel BC\), \(AD=18\), \(BC=12\). The diagonals meet at \(O\). Find \(AO:OC\).
Triangles \(AOD\) and \(COB\) are similar: the angles at \(O\) are vertical, and the other corresponding angles are equal because \(AD\parallel BC\). Therefore \(AO:OC=AD:BC=18:12=3:2\).
Example 9. A Tangent as a Source of Equal Angles
A tangent can unexpectedly lead to an isosceles triangle.
Problem. In triangle \(ABC\), the tangent to the circumcircle at \(A\) is parallel to \(BC\). Prove that \(AB=AC\).
The angle between the tangent and \(AB\) equals \(\angle ACB\). Since the tangent is parallel to \(BC\), this same angle equals \(\angle ABC\). Hence \(\angle ABC=\angle ACB\), so \(AB=AC\).
Example 10. Areas Determine Medians
Sometimes equality of areas shows where a median passes.
Problem. Point \(P\) lies inside triangle \(ABC\), and \(S_{PAB}=S_{PAC}=S_{PBC}\). Prove that \(P\) is the intersection point of the medians.
From \(S_{PAB}=S_{PAC}\), line \(AP\) passes through the midpoint of \(BC\). From \(S_{PAB}=S_{PBC}\), line \(BP\) passes through the midpoint of \(AC\). Therefore \(P\) is the intersection of two medians, hence the centroid of the triangle.