China Western Mathematical Olympiad

Solution:

As seen in the figure, let point $I$ be the center of the inscribed circle of $ABCD$. Since $H$ is the midpoint of $D_1{A}_1$ and lines $A{A}_1$, $A{D}_1$ are the two tangent lines through $A$ to the circle, then point $H$ lies on the line segment $AI$ and $AI\perp A_1{D}_1$. From $I{D}_1\perp A{D}_1$ and using the proportional theorem for similar triangles we get that $IH\cdot IA=I{D}_1^2=r^2$, where $r$ is the radius of the inscribed circle. In the same way, we get that $IE\cdot IB=r^2$. Then $IE\cdot IB=IH\cdot IA$, and that means points $A$, $H$, $E$, $B$ lie on one circle, so $\angle EHI=\angle ABE$.

Similarly, we obtain that $\angle IHG=\angle ADG$, $\angle IFE=\angle CBE$, $\angle IFG=\angle CDG$. Adding these four equations, we get that $\angle EHG+\angle EFG=\angle ABC+\angle ADC$, and that means $A$, $B$, $C$, $D$ lie on one circle if and only if $E$, $F$, $G$, $H$ lie on one circle.

Notice that quadrilateral $EFGH$ is a parallelogram as $E$, $F$, $G$, $H$ are the midpoints of the sides of quadrilateral $A_1{B}_1{C}_1{D}_1$ respectively. So $E$, $F$, $G$, $H$ lie on one circle if and only if $EFGH$ is a rectangle. This completes the proof.