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Iranian Mathematical Olympiad

Geometry Difficulty 4.8 AIME Prove it Iran

$\triangle ABC$ is an isosceles triangle with $AB = AC$ . Point $X$ is an arbitrary point on side $BC$ . Points $Y, Z$ are on the sides $AB, AC$ , respectively, such that $\angle BXY = \angle ZXC$ . A line parallel to $YZ$ and passing through $B$ cuts $XZ$ at $T$ . Prove that $AT$ bisects $\angle A$ .

Solution

Let us denote by $K$ the intersection point of lines $BT$ , $XY$ . Note that $\triangle XYB \sim \triangle XZC$ . Using the fact that $BT \parallel YZ$ , we get
$\frac{XK}{YK} = \frac{XT}{ZT}.$
Therefore, $K$ and $T$ are corresponding points in triangles $\triangle XYB$ and $\triangle XZC$ .
Which gives us
$\angle TCX = \angle KBX = \angle TBX,$
and $TB = TC$ . So, $T$ lies on the perpendicular bisector of $BC$ and since $AB = AC$ , $T$ also lies on the angle bisector of $\angle A$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.