Angles, Lines and Parallel Lines

Let the smaller angle be \(x\). Then the larger one is \(x+28^\circ\), and \(x+x+28^\circ=180^\circ\). Hence \(2x=152^\circ\), so \(x=76^\circ\). The other angle is \(104^\circ\).

Angles \(AOC\) and \(BOD\) are vertical, so \(\angle BOD=64^\circ\). Angle \(AOD\) is adjacent to \(AOC\), hence \(\angle AOD=180^\circ-64^\circ=116^\circ\). Angle \(BOC\) is vertical to \(AOD\), so \(\angle BOC=116^\circ\).

Let the angles be \(2x\), \(3x\), and \(4x\). Their sum is \(180^\circ\), hence \(9x=180^\circ\) and \(x=20^\circ\). The angles are \(40^\circ\), \(60^\circ\), and \(80^\circ\).

We have \(126^\circ=\angle A+\angle B\), so \(\angle B=126^\circ-49^\circ=77^\circ\). The interior angle at \(C\) is adjacent to the given exterior angle, hence \(\angle C=180^\circ-126^\circ=54^\circ\).

The alternate interior angle is \(117^\circ\). The same-side interior angle is \(180^\circ-117^\circ=63^\circ\).

Angles \(\angle CAB\) and \(\angle ABD\) are alternate interior angles for lines \(AC\), \(BD\) with transversal \(AB\). They are equal by the condition. By the parallelism criterion, \(AC\parallel BD\).

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. The sum of these angles is \(\frac{\alpha+\beta}{2}=90^\circ\). Therefore the angle between the bisectors is \(90^\circ\), so they are perpendicular.

Since \(l\parallel AC\), the two angles at line \(l\) are equal to angles \(A\) and \(C\) of the triangle, while the middle angle is \(\angle B\). Thus the triangle angles are in the ratio \(3:8:4\). Let them be \(3x\), \(8x\), and \(4x\). Then \(15x=180^\circ\), so \(x=12^\circ\). The angles are \(36^\circ\), \(96^\circ\), and \(48^\circ\).

Let two angles be \(x\) and \(y\), and the third be \(x+y\). The angle sum gives \(x+y+(x+y)=180^\circ\). Therefore \(2(x+y)=180^\circ\), so \(x+y=90^\circ\). The third angle is \(90^\circ\), hence the triangle is right-angled.

We have \(\angle A=180^\circ-42^\circ-74^\circ=64^\circ\). Since \(AD\) is the angle bisector, \(\angle BAD=32^\circ\). In triangle \(ABD\), \(\angle BDA=180^\circ-42^\circ-32^\circ=106^\circ\). Hence \(\angle ADC=180^\circ-106^\circ=74^\circ\).

Since \(AB=AC\), the base angles are equal: \(\angle B=\angle C=72^\circ\). Then \(\angle A=180^\circ-72^\circ-72^\circ=36^\circ\). The angle bisector \(BD\) gives \(\angle ABD=\angle DBC=36^\circ\). In triangle \(ABD\), the angles are \(36^\circ\), \(36^\circ\), \(108^\circ\). In triangle \(BCD\), \(\angle DBC=36^\circ\), \(\angle BCD=72^\circ\), hence \(\angle BDC=72^\circ\).

First, \(\angle A=180^\circ-54^\circ-62^\circ=64^\circ\). Since \(DE\parallel BC\), we have \(\angle ADE=\angle C=62^\circ\) and \(\angle AED=\angle B=54^\circ\). Therefore the angles of triangle \(ADE\) are \(64^\circ\), \(62^\circ\), and \(54^\circ\).

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 \(90^\circ-\beta\). Thus the angle between \(AH\) and \(AL\) equals \(\frac{180^\circ-\beta-\gamma}{2}-(90^\circ-\beta)=\frac{\beta-\gamma}{2}\). If \(\gamma>\beta\), the same argument gives \(\frac{\gamma-\beta}{2}\).

In triangle \(BIC\), \(\angle IBC=\frac{B}{2}\) and \(\angle ICB=\frac{C}{2}\). Hence \(\angle BIC=180^\circ-\frac{B+C}{2}\). Since \(B+C=180^\circ-A\), we get \(\angle BIC=180^\circ-\frac{180^\circ-A}{2}=90^\circ+\frac{A}{2}\). Thus \(124^\circ=90^\circ+\frac{A}{2}\), so \(A=68^\circ\).

Let the exterior bisector make an angle \(x\) with side \(AB\). Since it is parallel to \(BC\), this angle equals \(\angle B\). Since it bisects the exterior angle, the second angle between it and line \(AC\) is also \(x\). By parallelism, it equals \(\angle C\). Therefore \(\angle B=\angle C\). The triangle is isosceles, so \(AB=AC\).

In triangle \(ABC\), \(\angle C=180^\circ-60^\circ-50^\circ=70^\circ\). Since \(D\) lies on \(AB\), angle \(CBD\) equals \(\angle CBA=50^\circ\). In triangle \(BCD\), sides \(CD\) and \(BD\) are equal, so the base angles at \(BC\) are equal: \(\angle BCD=\angle CBD=50^\circ\). Hence \(\angle ACD=\angle ACB-\angle BCD=70^\circ-50^\circ=20^\circ\).

Since \(AB=AC\), the base angles are equal: \(\angle B=\angle C=\frac{180^\circ-36^\circ}{2}=72^\circ\). In triangle \(BCD\), sides \(BD\) and \(BC\) are equal, so \(\angle BDC=\angle BCD\). Since \(D\) lies on \(AC\), \(\angle BCD=72^\circ\), hence \(\angle BDC=72^\circ\). Therefore \(\angle DBC=180^\circ-72^\circ-72^\circ=36^\circ\). Thus \(\angle ABD=\angle ABC-\angle DBC=72^\circ-36^\circ=36^\circ\).

Angles \(\angle BAC\) and \(\angle DCA\) are alternate interior angles for lines \(AB\) and \(CD\) with transversal \(AC\). They are equal, so \(AB\parallel CD\). Similarly, angles \(\angle BCA\) and \(\angle DAC\) are alternate interior angles for lines \(BC\) and \(AD\) with transversal \(AC\). They are equal, hence \(BC\parallel AD\).

Since \(DE\parallel AB\), the angle between \(DE\) and \(AD\) equals \(\angle BAD\). Since \(DF\parallel AC\), the angle between \(AD\) and \(DF\) equals \(\angle DAC\). By the condition, \(\angle BAD=\angle DAC\). Hence \(AD\) divides \(\angle EDF\) into two equal angles, so it is its angle bisector.

Since \(DE\parallel AB\), the angle between \(DE\) and \(AD\) equals \(\angle BAD\). Since \(DF\parallel AC\), the angle between \(AD\) and \(DF\) equals \(\angle DAC\). By the condition, \(AD\) bisects \(\angle EDF\), so these two angles at \(D\) are equal. Therefore \(\angle BAD=\angle DAC\), meaning that \(AD\) bisects angle \(A\).