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Romanian Mathematical Olympiad - District Round

Geometry Difficulty 6.3 National Olympiad Prove it Romania

The triangle $ABC$ is isosceles, with $AB = AC$ and $\angle ABC = 72^\circ$ . The point $D$ is taken on the line $BC$ , so that $C$ is on the segment $BD$ and $CD = AB$ .

a) Prove that $AC$ is the bisector of the angle $\angle BAD$ .

b) The point $E$ is taken on the parallel to $AB$ through $D$ , on the same side of $BD$ as $A$ , so that $DE = DB$ . Let $F$ be the common point of the lines $AD$ and $BE$ . Prove that the lines $AC$ and $AE$ are perpendicular and $AF = FC = BC$ .

Solution

a) In the triangle $ABC$ , $\angle BAC = 36^\circ$ . From the isosceles triangle $ACD$ , with $\angle ACD = 180^\circ - 36^\circ = 144^\circ$ , it follows that $\angle CAD = \angle ADC = 36^\circ$ , hence $AC$ is the bisector of the angle $\angle BAD$ .

b) From $\angle ABD = \angle BAD = 72^\circ$ it follows that the triangle $ABD$ is isosceles, hence $AD = BD$ . Since $DE = DB$ , the triangle $ADE$ is isosceles, therefore $\angle DAE = (180^\circ - \angle ADE) : 2 = (180^\circ - 72^\circ) : 2 = 54^\circ$ . So, $\angle CAE = \angle CAD + \angle DAE = 36^\circ + 54^\circ = 90^\circ$ .
From $\angle AFB = 180^\circ - \angle BAF - \angle ABF = 72^\circ$ it follows $\triangle BAC \equiv \triangle ABF$ (S.A.S.), whence $BC = AF$ . Now $BD = AD$ yields $CD = FD$ . This leads to $\triangle BAC \equiv \triangle CDF$ (S.A.S.), hence $BC = FC$ .

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