NZMO Round One

Geometry Difficulty 4.5 AIME Prove it New Zealand

Problem:

Let $ABC$ be an isosceles triangle with $AB = AC$ . Point $D$ lies on side $AC$ such that $BD$ is the angle bisector of $\angle ABC$ . Point $E$ lies on side $BC$ between $B$ and $C$ such that $BE = CD$ . Prove that $DE$ is parallel to $AB$ .

Solution

Solution:

Let $E'$ be the point on line $BC$ such that $DE'$ is parallel to $AB$ . We know that $E'$ lies between $B$ and $C$ because $D$ lies between $A$ and $C$ . So it suffices for us to prove that $BE' = CD$ .

$\angle ACB = \angle CBA = \angle CE'D$ .

Therefore triangle $CDE'$ is isosceles with $CD = DE'$ .

$\angle E'BD = \angle DBA = \angle ADE'$ .

Therefore triangle $BE'D$ is isosceles with $BE' = DE'$ . Hence $CD = DE' = BE'$ as required.

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