Mixed Problems I

In triangle \(ABC\), \(\angle C=180^\circ-46^\circ-71^\circ=63^\circ\). Since \(DE\parallel BC\), angle \(\angle ADE\) equals \(\angle ACB\). Therefore \(\angle ADE=63^\circ\).

We have \(AB=AC\), \(BM=MC\), and \(AM\) is common. Therefore \(\triangle ABM\cong\triangle ACM\). Hence \(\angle AMB=\angle AMC\). These angles are supplementary, so each is \(90^\circ\), and \(AM\perp BC\).

From \(DE\parallel BC\), \(\triangle ADE\sim\triangle ABC\). Since \(AD:AB=4:5\), we have \(DE:BC=4:5\). Therefore \(DE=25\cdot\frac{4}{5}=20\).

Triangles \(ABD\) and \(ABC\) have the common height from \(A\) to \(BC\). Since \(BD:BC=2:7\), we get \(S_{ABD}=\frac{2}{7}\cdot84=24\).

Angles \(\angle ABC\) and \(\angle ADC\) are inscribed and stand on the same chord \(AC\). With the given position, they are equal. Therefore \(\angle ADC=39^\circ\).

Since \(BD\perp AC\), we have \(\angle BDC=90^\circ\). Since \(CE\perp AB\), we have \(\angle BEC=90^\circ\). Hence points \(D\) and \(E\) lie on the circle with diameter \(BC\). Therefore \(B,C,D,E\) are cyclic.

Point \(M\) is the midpoint of \(BC\) and, by construction, the midpoint of \(AD\). Hence the diagonals of quadrilateral \(ABDC\) are bisected by point \(M\). Therefore \(ABDC\) is a parallelogram, so \(AB\parallel CD\) and \(AC\parallel BD\).

Since \(DE\parallel AC\), \(\triangle BDE\sim\triangle BAC\). From \(BD:BA=1:2\), we get \(BE:BC=1:2\). Hence \(BE=EC\), so \(E\) is the midpoint of \(BC\).

Draw \(AC\). In triangle \(ABC\), segment \(MN\) is a midline, so \(MN\parallel AC\). In triangle \(CDA\), segment \(PQ\) is a midline, so \(PQ\parallel AC\). Therefore \(MN\parallel PQ\). Similarly, using diagonal \(BD\), we get \(NP\parallel MQ\). Hence \(MNPQ\) is a parallelogram.

Triangles \(AOD\) and \(COB\) are similar: the angles at \(O\) are vertical, and the other corresponding angles are equal because \(AD\parallel BC\). Therefore \(AO:OC=DO:OB=AD:BC=21:14=3:2\).

The angle between the tangent and chord \(AB\) equals \(\angle ACB\). Since the tangent is parallel to \(BC\), the same angle equals \(\angle ABC\). Hence \(\angle ABC=\angle ACB\), so \(AB=AC\).

Since \(P\) lies on median \(AM\), line \(AP\) passes through the midpoint of \(BC\). Triangles \(PAB\) and \(PAC\) have common base \(AP\), and points \(B\) and \(C\) are equally distant from line \(AP\). Therefore the areas are equal.

Triangles \(ABD\) and \(ACD\) have common base \(AD\). The heights from \(B\) and \(C\) to \(AD\) are equal because \(BC\parallel AD\). Hence \(S_{ABD}=S_{ACD}\). Subtracting the common area \(S_{AOD}\), we get \(S_{AOB}=S_{COD}\).

Since \(AB\parallel CD\), \(\angle ABD=\angle BDC\). In the cyclic quadrilateral, \(\angle BDC=\angle BAC\), because both angles stand on chord \(BC\). Hence \(\angle ABD=\angle BAC\). These angles stand on chords \(AD\) and \(BC\), so the chords are equal: \(AD=BC\).

Triangles \(BCE\) and \(BCA\) have common base \(BC\). Since \(D\) lies on \(BC\), the height of point \(E\) to \(BC\) is \(ED:AD=2:5\) of the height of point \(A\). Therefore \(S_{BCE}=\frac{2}{5}\cdot120=48\).

Since \(BD\perp AC\), \(\angle ADH=90^\circ\). Since \(CE\perp AB\), \(\angle AEH=90^\circ\). Hence \(A,D,H,E\) lie on the circle with diameter \(AH\). Also, \(HD\perp AC\) and \(HE\perp AB\), so the obtuse angle between \(HD\) and \(HE\) equals \(180^\circ-\angle A\). Therefore \(\angle DHE=180^\circ-\angle A\).

Let the distances from \(P\) to lines \(AB\) and \(CD\) be \(h_1\) and \(h_2\). Then \(h_1+h_2\) equals the height of the parallelogram to base \(AB\). Since \(AB=CD\), we have \(S_{PAB}+S_{PCD}=\frac{1}{2}ABh_1+\frac{1}{2}ABh_2=\frac{1}{2}S_{ABCD}\).

Since \(AC\) bisects \(\angle BAD\), \(\angle BAC=\angle CAD\). Angle \(\angle BAC\) stands on chord \(BC\), and \(\angle CAD\) stands on chord \(CD\). Equal inscribed angles stand on equal chords, hence \(BC=CD\).

Let the angle between the diagonals be \(\varphi\). Then \(S_{AOB}=\frac{1}{2}AO\cdot BO\sin\varphi\), \(S_{COD}=\frac{1}{2}CO\cdot DO\sin\varphi\). Similarly, \(S_{BOC}=\frac{1}{2}BO\cdot CO\sin\varphi\), \(S_{DOA}=\frac{1}{2}DO\cdot AO\sin\varphi\). The products are equal.

Since \(DE\parallel AC\), \(\triangle BDE\sim\triangle BCA\). The similarity ratio is \(BD:BC=2:5\), so \(S_{BDE}:S_{ABC}=4:25\). The remaining part \(ADEC\) has area \(21\) parts out of \(25\). Hence \(S_{BDE}:S_{ADEC}=4:21\).

Draw \(AC\). Since \(AB\parallel CD\), \(\angle BAC=\angle ACD\). Also \(AB=CD\), and \(AC\) is common. Therefore \(\triangle BAC\cong\triangle DCA\). Hence \(\angle BCA=\angle CAD\), so \(BC\parallel AD\). Together with \(AB\parallel CD\), this gives a parallelogram.

Angles \(\angle CDE\) and \(\angle CBE\) stand on the same segment \(CE\). If two points \(D\) and \(B\) see segment \(CE\) under equal angles, then they lie on one circle with \(C\) and \(E\). Therefore \(B,C,D,E\) are cyclic.

Since \(M\) is the midpoint of \(BC\), \(BM=CM\). By condition \(AM=BM\), hence \(AM=BM=CM\). Therefore points \(A,B,C\) lie on the circle with centre \(M\), and \(BC\) is a diameter. Angle \(\angle BAC\) stands on the diameter, so it is right.

Triangles \(PAB\) and \(PAC\) have common base \(AP\). Equality of areas means that the distances from \(B\) and \(C\) to line \(AP\) are equal. Since \(B\) and \(C\) are on opposite sides of \(AP\), line \(AP\) passes through the midpoint of \(BC\).

Let \(AD=a\), \(BC=b\). From \(AOD\sim COB\), we get \(DO:OB=a:b\). In triangle \(DBA\), segment \(OX\parallel AD\), so \(OX=\frac{ab}{a+b}\). Similarly, in triangle \(DBC\), \(OY=\frac{ab}{a+b}\). Hence \(XY=\frac{2ab}{a+b}\). For \(a=12\), \(b=6\), we get \(XY=8\).

Through areas, \(\frac{BD}{DC}=\frac{S_{ABP}}{S_{ACP}}\), \(\frac{CE}{EA}=\frac{S_{BCP}}{S_{ABP}}\), and \(\frac{AF}{FB}=\frac{S_{ACP}}{S_{BCP}}\). Multiplying the three equalities, all areas cancel, and the result is \(1\).

By the area form of Ceva, \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1\). Thus \(\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{AF}{FB}=1\), so \(\frac{1}{2}\cdot\frac{AF}{FB}=1\). Therefore \(AF:FB=2:1\).

Since \(G\) lies on the median from \(A\), \(S_{GAB}=S_{GAC}\). Since \(G\) lies on the median from \(B\), \(S_{GAB}=S_{GBC}\). Hence \(S_{GAC}=S_{GBC}\). These triangles have common base \(GC\), so points \(A\) and \(B\) are equally distant from line \(GC\). Therefore \(GC\) passes through the midpoint of \(AB\), that is, \(AF=FB\).

Let \(O\) be the centre of the circle. Radii \(OB\) and \(OC\) are perpendicular to the tangents, so in quadrilateral \(BOCT\), the angles at \(B\) and \(C\) are right. The central angle \(\angle BOC=2\angle BAC\). Hence \(\angle BTC=360^\circ-90^\circ-90^\circ-2\angle BAC=180^\circ-2\angle BAC\).

We have \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=\frac{2}{3}\cdot\frac{3}{5}\cdot\frac{5}{2}=1\). By the converse form of the area version of Ceva's theorem, it follows that lines \(AD\), \(BE\), \(CF\) meet at one point.