BxMO Team Selection Test, March 2020

Let $K$ be the midpoint of $D{D}_1$, and let $L$ be the midpoint of $D{D}_2$. 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 $D{D}_1{D}_2$, hence $\angle DLK=\angle D{D}_2{D}_1$. We conclude that $\angle D{D}_2{D}_1=90^{\circ }-\angle ABC$.

Because $AC\perp D{D}_2$ and $D_2{E}_2\parallel AC$, we have $\angle D{D}_2{E}_2=90^{\circ }$. Hence, $\angle D_1{D}_2{E}_2=\angle D_1{D}_{2}D+\angle D{D}_2{E}_2=90^{\circ }-\angle ABC+90^{\circ }=180^{\circ }-\angle ABC$. On the other hand, as $D_1{E}_1\parallel AB$, we have $\angle D_1{E}_1{E}_2=\angle ABC$, hence we get $\angle D_1{D}_2{E}_2=180^{\circ }-\angle D_1{E}_1{E}_2$. We conclude that $D_1{E}_1{E}_2{D}_2$ is a cyclic quadrilateral.

Let $M$ be the point such that $AM$ is a diameter of the circumcircle of $△ABC$. Thales' theorem yields $\angle ACM=90^{\circ }$. Hence, $CM\perp AC$, which yields $CM\perp D_2{E}_2$ and $CM\parallel D{D}_2$. Moreover, $L$ is the midpoint of $D{D}_2$ and $LC\parallel D_2{E}_2$, hence $LC$ is a midsegment in triangle $D{D}_2{E}_2$. This means that $C$ is the midpoint of $D{E}_2$. Because $CM\parallel D{D}_2$, we get that $CM$ is also a midsegment, hence $CM$ intersects $D_2{E}_2$ in the middle. As $CM\perp D_2{E}_2$, the line $CM$ is the perpendicular bisector of $D_2{E}_2$. Analogously, we get that $BM$ is the perpendicular bisector of $D_1{E}_1$. 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. $\square$