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Belarusian Mathematical Olympiad

Geometry Difficulty 6.8 National Olympiad Prove it Belarus

The altitudes $AA_1$ , $BB_1$ and $CC_1$ are drawn in the acute triangle $ABC$ . The bisector of the angle $AA_1C$ intersects the segments $CC_1$ and $CA$ at $E$ and $D$ respectively. The bisector of the angle $AA_1B$ intersects the segments $BB_1$ and $BA$ at $F$ and $G$ respectively. The circumcircles of the triangles $FA_1D$ and $EA_1G$ intersect at $A_1$ and $X$ .
Prove that $\angle BXC = 90^\circ$ .

Solution

Since $\angle BB_1C = \angle CC_1B = 90^\circ$ , it is sufficient to prove that the points $B$ , $C_1$ , $X$ and $B_1$ lie on a circle. Since $A_1D$ and $A_1G$ are the bisectors of the angles $\angle AA_1C$ and $\angle AA_1B$ respectively, $\angle GA_1D = 90^\circ$ . Since $\angle FB_1D = \angle FA_1D = 90^\circ$ , the point $B_1$ lies on the circumcircle of the triangle $FA_1D$ . Similarly, the point $C_1$ lies on the circumcircle of the triangle $EA_1G$ .

Therefore $\angle C_1XA_1 = \angle BGA_1 = 135^\circ - \angle ABC$ and $\angle B_1XA_1 = \angle CDA_1 = 135^\circ - \angle ACB$ . Hence $\angle C_1XB_1 = 270^\circ - \angle ABC - \angle ACB = 90^\circ + \angle BAC$ . Together with $\angle B_1BA = 90^\circ - \angle BAC$ , this leads to $\angle B_1BC_1 + \angle C_1XB_1 = 180^\circ$ , which means that the quadrilateral $BC_1XB_1$ is cyclic.

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