Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.5 National olympiad Prove it Belarus

Let $M$ be the midpoint of the hypotenuse $AB$ of the right triangle $ABC$ . Point $P$ is chosen on the cathetus $CB$ so that $CP : PB = 1 : 2$ . The straight line passing through $B$ meets the segments $AC, AP,$ and $PM$ at points $X, Y, \text{ and } Z$ , respectively.
Prove that the bisector of the angle $PZY$ passes through point $C$ if and only if the bisector of the angle $PYX$ also passes through $C$ .

Solution

Let $D$ be symmetric to $A$ with respect to the vertex $C$ (see the Fig.). The segment $BC$ is the altitude and the median in the triangle $ABD$ so the triangle $ABD$ is isosceles. Likewise, the triangle $ADP$ is isosceles. Let $P'$ be the intersection point of the medians $DM$ and $BC$ of the triangle $ABD$ . Since the point of intersection divides the medians in the ratio $2:1$ , points $P$ and $P'$ coincide. Since $C$ belongs to the bisector of the angle $\angle DPA$ , we see that $C$ is equidistant to the lines $DP$ and $PA$ .
If $C$ belongs to the bisection of the angle $\angle PZY$ , then $C$ is equidistant to the lines $PZ$ and $ZY$ , therefore, $C$ is equidistant to the lines $YP$ and $YX$ , i.e. $C$ belongs to the bisection of the angle $\angle PYX$ .
In the same way, one can prove the converse proposition.

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