Euler Olympiad

Geometry Difficulty 5.6 AIME, harder Prove it Russia

In triangle $ABC$ , sides $AB$ and $BC$ have equal lengths. Point $D$ inside the triangle is chosen so that $\angle ADC = 2\angle ABC$ . Prove that the distance from point $B$ to the external bisector line of angle $\angle ADC$ is twice smaller than $AD + DC$ . (S. Berlov)

Solution

Let $\ell$ be the external angle bisector of the angles adjacent to $\angle ADC$ , and let $K$ be the projection of $B$ onto $\ell$ . Let points $B'$ and $C'$ be the reflections of $B$ and $C$ with respect to $\ell$ , respectively. Then $BB' = 2BK$ —that is, twice the distance from $B$ to $\ell$ . Moreover, point $D$ lies on segment $AC'$ (since lines $DA$ and $DC$ are symmetric with respect to $\ell$ ), and $AC' = AD + DC' = AD + DC$ .

Furthermore, by the same symmetry, we have $\angle AC'B' = \angle DC'B' = \angle DCB$ , $\angle BB'C' = \angle B'BC$ .

Let segments $BB'$ and $AC'$ intersect at point $O$ . From the right triangle $OKD$ , we get $\angle BOC' = \angle KOD = 90^\circ - \angle KDO = \frac{1}{2}(180^\circ - \angle CDC') = \frac{1}{2}\angle ADC = \angle ABC$ . Therefore, $\angle ABB' = \angle ABC - \angle B'BC = \angle BOC' - \angle OB'C' = \angle OC'B'$ . Similarly, $\angle BAO = \angle BOC' - \angle ABO = \angle ABC - \angle ABO = \angle B'BC = \angle BB'C'$ .

Since segments $BC$ and $B'C'$ are symmetric, we have $B'C' = BC = AB$ . Thus, triangles $ABO$ and $B'C'O$ are congruent by a side and two adjacent angles. Hence $BB' = BO + OB' = C'O + OA = AC' = AD + DC$ , as required.

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