Given is an isosceles trapezoid $ABCD$, such that $AD$ and $BC$ are bases and $AD=2AB$, and it is inscribed in a circle $c$. Points $E$ and $F$ are selected on a circle $c$ so that $AC$ || $DE$ and $BD$ || $AF$. The line $BE$ intersects lines $AC$ and $AF$ at points $X$ and $Y$, respectively. Prove that the circumcircles of triangles $BCX$ and $EFY$ are tangent to each other.
Problem
Source: St Petersburg 2021 9.5
Tags: geometry
Tintarn
13.01.2022 14:21
We show that both circles pass through $M$, the midpoint of $AD$. Then automatically (since everything is symmetric over the perpendicular bisector of $AD$), the line $AD$ will be a common tangent of both, and the claim follows.
Let $Z$ be the intersection of $BD$ with $(BCX)$. So $Z$ lies symmetric to $X$. Now
\[\angle ZBX=\angle DBE=\angle DAE=\angle MAE=\frac{1}{2} \angle MBX,\]where we used that $B,M,E$ are on a common circle centered at $A$.
Hence $BM$ is the angular bisector of $\angle ZBX$, and since it is also on the perpendicular bisector, it is on the circumcircle, as desired.
The claim for $(EFY)$ follows in just the same way.
deepakies
28.02.2022 17:28
@above correction it will rather be \[\angle ZBX=\angle DBE=\angle DAE=\angle MAE=2 \angle MBX,\]Elaborating upon the claim for $(EFY)$: Similar to $Z$ let $W$ be the intersection of $DE$ with $CF$ making it symmetric to $Y$. Angle chasing: $$\angle WEY= \angle DEB=\angle DAB=\angle MAB=2\angle MEB=2\angle MEY $$. Thus, $M$ lies on both circles and we are done.