Competition problem — Geometry — Maths Olympiad Prep

Library / /10 of 189

Geometry Difficulty 4.8 AIME Prove it Ukraine

There is isosceles obtuse triangle $ABC$ with vertex in point $B$ given. Perpendicular bisector to side $BC$ intersects lines $AC$ and $AB$ in points $K$ and $M$ respectively. Prove, that point, symmetric to point $A$ with respect to line $BK$ , is on line $CM$ .

(Anton Trigub)

Solution

Let $A_1$ be point, symmetric to $A$ with respect to $BK$ (Fig. 37). At first, $\angle BA_1K = \angle BAK = \angle BCK$ , thus quadrilateral $BA_1CK$ is cyclic. Then,
$\angle A_1CB = \angle A_1KB = \angle AKB = 2\angle BCA = \angle MBC = \angle MCB,$
So we get, that points $C$ , $A_1$ , $M$ are on one line.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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