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Geometry Difficulty 8.3 Shortlist Prove it Saudi Arabia

Let $ABC$ be an acute triangle, $M$ be the midpoint of $BC$ and $P$ be a point on line segment $AM$ . Lines $BP$ and $CP$ meet the circumcircle of $ABC$ again at $X$ and $Y$ , respectively, and sides $AC$ at $D$ and $AB$ at $E$ , respectively. Prove that the circumcircles of $AXD$ and $AYE$ have a common point $T \neq A$ on line $AM$ .

Solution

By applying Ceva to the concurrent cevians $AM$ , $BD$ and $CE$ , we obtain
$\frac{AE}{EB} = \frac{AD}{DC} \cdot \frac{CM}{MB} = \frac{AD}{DC}.$
We deduce from Thales' theorem that segments $ED$ and $BC$ are parallel.

Since quadrilateral $BCXY$ is cyclic, we have $\angle BXY = \angle BCY = \angle CED$ . We deduce that quadrilateral $EDXY$ is cyclic and therefore
$PD \cdot PX = PE \cdot PY$
Hence, point $P$ lies on the radical axis of circumcircles of triangles $AXD$ and $AYE$ which passes through $A$ .

If these two circles are tangent to $AP$ at $A$ then
$\angle MAC = \angle AXB = \angle ACB, \text{ and } \angle BAM = \angle CYA = \angle CBA,$
which implies that triangle $ABC$ is a right triangle at $A$ and this is a contradiction.

We conclude that the two circles intersect in a second point $T \neq A$ on line $AM$ .

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