Japan Junior Mathematical Olympiad

Take a point $E^{\prime }$ on the opposite side of the line $BC$ from $A$ in such a way that the triangle $BC{E}^{\prime }$ becomes congruent with the triangle $ADE$. Since $ABCD$ is a square, we have $AC=BD$ and $\angle DAC=\angle CBD$. We also have $AE=B{E}^{\prime }$ and $\angle EAD=\angle E^{\prime }BC$. Therefore, we get $\angle EAC=\angle EAD+\angle DAC=\angle E^{\prime }BC+\angle CBD=\angle E^{\prime }BD$, and we conclude that the triangles $ACE$ and $BD{E}^{\prime }$ are congruent. Then, we see that $\angle BED+\angle B{E}^{\prime }D=\angle BED+\angle AEC=180^{\circ }$, which implies that the 4 points $B$, $D$, $E$, $E^{\prime }$ lie on the circumference of a circle.

Furthermore, from $$\begin{array}{rl}\angle BEC+\angle B{E}^{\prime }C& =\angle BEC+\angle AED\\ & =(\angle BED-\angle CED)+(\angle AEC+\angle CED)\\ & =\angle BED+\angle AEC=180^{\circ },\end{array}$$ we see that the 4 points $B$, $C$, $E$, $E^{\prime }$ also lie on the circumference of a circle. Since the circle going through 3 points $B$, $E$, $E^{\prime }$ is unique (which is the circumcircle of the triangle $BE{E}^{\prime }$), and since the points $C$ and $D$ lie on this circle, we conclude that the circle is the circumcircle of the square $ABCD$ and that the point $E$ lies on it.

Alternate Solution: Let us denote by $\Gamma$ the circumcircle of the square $ABCD$ and let $O$ be the center of $\Gamma$. Denote by $F$ the point of intersection of the half line $OE$ and $\Gamma$. It is enough to show that if $F$ and $E$ do not coincide, then $\angle AEC+\angle BED\ne 180^{\circ }$ must hold.

We note first that $\angle AFC=\angle BFD=90^{\circ }$ holds since the line segments $AC$, $BD$ are the diameters of $\Gamma$.

Assume that $E$ and $F$ are distinct and the 3 points $O$, $E$, $F$ lie on the line in this order. Then, we have $\angle AEC=\angle AFC-\angle EAF-\angle ECF<90^{\circ }$. Similarly, we get $\angle BED<90^{\circ }$ and therefore, $\angle AEC+\angle BED<180^{\circ }$.

If, on the other hand, the points lie on the line in the order of $O$, $E$, $F$, we get $\angle AEC=\angle AFC+\angle EAF+\angle ECF>90^{\circ }$, and similarly, $\angle BED>90^{\circ }$ and therefore, $\angle AEC+\angle BED>180^{\circ }$. Our claim is thus established.