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Belarusian Mathematical Olympiad

Geometry Difficulty 6.5 National Olympiad Prove it Belarus

A point $P$ is chosen in the interior of the side $BC$ of the triangle $ABC$ . The points $D$ and $E$ are symmetric to $P$ with respect to the vertices $B$ and $C$ respectively. The circumcircles of the triangles $ABE$ and $ACD$ intersect at the points $A$ and $X$ . The ray $AB$ intersects the segment $XD$ at the point $C_1$ and the ray $AC$ intersects the segment $XE$ at the point $B_1$ .
Prove that the lines $BC$ and $B_1C_1$ are parallel.

Solution

First we will prove that point $X$ lies on the line $AP$ . Let the line $AP$ intersect the circumcircle of the triangle $ACD$ at points $A$ and $Y$ .

Since $PC \cdot PD = 2PB \cdot PC = PB \cdot PE$ , the equality $AP \cdot PY = PC \cdot DP$ implies $AP \cdot PY = PB \cdot PE$ , and therefore the points $A$ , $B$ , $Y$ and $E$ are concyclic, consequently $X \equiv Y$ .

From the circumcircles of $ACXD$ and $ABXE$ we obtain $\angle BAP = \angle BEX$ and $\angle CAP = \angle CDX$ .

The quadrilateral $AC_1XB_1$ is cyclic, since
$\angle DXE = 180^\circ - \angle CDX - \angle BEX = 180^\circ - \angle BAC.$
Therefore, $\angle C_1B_1X = \angle C_1AX$ , whence $\angle C_1B_1X = \angle BEX$ and $B_1C_1 \parallel DE$ .

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