Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.7 AIME Prove it Russia

A point $D$ is chosen on side $BC$ of an acute triangle $ABC$ so that $AB = AD$ . The circumcircle of triangle $ABD$ intersects segment $AC$ at points $A$ and $K$ . Line $DK$ intersects the line perpendicular to $AC$ and passing through $B$ at point $L$ . Prove that $CL = BC$ . (I. Bogdanov)

Solution

Так как $AB = AD$ , имеем $\angle ADB = \angle ABD$ . Поскольку четырехугольник $ABDK$ вписан, $\angle AKB = \angle ADB$ и $\angle ABD = 180^\circ - \angle AKD = \angle LKA$ . Таким образом, в треугольнике $BKL$ высота $KA$ является биссектрисой, а значит, и медианой; тогда точки $L$ и $B$ симметричны относительно $AC$ , поэтому отрезки $CL$ и $CB$ также симметричны. Значит, их длины равны.

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