BxMO Team Selection Test, March, 2020

Let $K$ be the midpoint of $DD_1$, and let $L$ be the midpoint of $DD_2$. Then $K$ lies on $AB$ and $L$ lies on $AC$. Because $\angle AKD = 90^\circ = \angle ALD$, the quadrilateral $AKDL$ is cyclic. Hence, $\angle DLK = \angle DAK = \angle DAB = 90^\circ - \angle ABC$. Moreover, $KL$ is a midsegment in triangle $DD_1D_2$, hence $\angle DLK = \angle DD_2D_1$. We conclude that $\angle DD_2D_1 = 90^\circ - \angle ABC$.

Because $AC \perp DD_2$ and $D_2E_2 \parallel AC$, we have $\angle DD_2E_2 = 90^\circ$. Hence, $\angle D_1D_2E_2 = \angle D_1D_2D + \angle DD_2E_2 = 90^\circ - \angle ABC + 90^\circ = 180^\circ - \angle ABC$. On the other hand, as $D_1E_1 \parallel AB$, we have $\angle D_1E_1E_2 = \angle ABC$, hence we get $\angle D_1D_2E_2 = 180^\circ - \angle D_1E_1E_2$. We conclude that $D_1E_1E_2D_2$ is a cyclic quadrilateral.

Let $M$ be the point such that $AM$ is a diameter of the circumcircle of $\triangle ABC$. Thales' theorem yields $\angle ACM = 90^\circ$. Hence, $CM \perp AC$, which yields $CM \perp D_2E_2$ and $CM \parallel DD_2$. Moreover, $L$ is the midpoint of $DD_2$ and $LC \parallel D_2E_2$, hence $LC$ is a midsegment in triangle $DD_2E_2$. This means that $C$ is the midpoint of $DE_2$. Because $CM \parallel DD_2$, we get that $CM$ is also a midsegment, hence $CM$ intersects $D_2E_2$ in the middle. As $CM \perp D_2E_2$, the line $CM$ is the perpendicular bisector of $D_2E_2$. Analogously, we get that $BM$ is the perpendicular bisector of $D_1E_1$. Hence, $M$ is the intersection point of the perpendicular bisectors of two of the chords of the circle through $D_1, D_2, E_1$, and $E_2$. Hence, $M$ is the centre of this circle. $\square$