Selection Examinations for the IMO

1st solution. Since $ACD$ is a right triangle, we have $|AC|^2 - |CD|^2 = |AD|^2$. The diagonals of the trapezoid intersect at a right angle, so $\angle DCA = \frac{\pi}{2} - \angle BDC = \angle ADB$. This, together with $\angle ADC = \frac{\pi}{2} = \angle BAD$, implies that the triangles $ACD$ and $BDA$ are similar. Thus, $\frac{|AD|}{|CD|} = \frac{|AB|}{|AD|}$ or, equivalently, $|AD|^2 = |AB| \cdot |CD|$. This implies
$$\frac{|AC| \cdot |CD|}{|AC|^2 - |CD|^2} = \frac{|AC| \cdot |CD|}{|AD|^2} = \frac{|AC| \cdot |CD|}{|AB| \cdot |CD|} = \frac{|AC|}{|AB|}.$$
Since $\angle BAD + \angle DEB = \frac{\pi}{2} + \frac{\pi}{2} = \pi$. $ABED$ is a cyclic quadrilateral. So, $\angle AEB = \angle ADB$. Because of the right angles we have $\angle BAC = \frac{\pi}{2} - \angle CAD = \angle ADB$. The triangles $ABE$ and $CBA$ are similar, since $\angle AEB = \angle BAC$ and $\angle EBA = \angle ABC$. Hence, $\frac{|AE|}{|BE|} = \frac{|CA|}{|AB|}$ and this implies
$$\frac{|AC| \cdot |CD|}{|AC|^2 - |CD|^2} = \frac{|AC|}{|AB|} = \frac{|AE|}{|BE|}.$$

2nd solution. Since $ACD$ is a right triangle, we have $|AC|^2 - |CD|^2 = |AD|^2$. Thus, it suffices to show that
$$\frac{|AE|}{|BE|} = \frac{|AC| \cdot |CD|}{|AD|^2}.$$
Let $T$ be the intersection of the diagonals. Obviously, $\angle TAD = \angle ABT$, so the triangles $ACD$ and $BDA$ are similar. Hence, $\frac{|AD|}{|CD|} = \frac{|AB|}{|AD|}$ or, equivalently, $|AD|^2 = |AB| \cdot |CD|$. We have $\angle BAD + \angle DEB = \frac{\pi}{2} + \frac{\pi}{2} = \pi$, so the quadrilateral $ABED$ is cyclic and $\angle EAD = \angle EBD$. Similarly, we have $\angle CTD + \angle DEC = \pi$, so the quadrilateral $CEDT$ is cyclic and $\angle CDE = \angle CTE$. Since the triangles $AED$ and $BET$ are similar, we have $\frac{|AE|}{|BE|} = \frac{|AD|}{|BT|}$. The triangles $ADC$ and $BTA$ are similar, so $\frac{|AD|}{|BT|} = \frac{|AC|}{|AB|}$. Hence,
$$\frac{|AE|}{|BE|} = \frac{|AC|}{|AB|} = \frac{|AC| \cdot |CD|}{|AB| \cdot |CD|} = \frac{|AC| \cdot |CD|}{|AD|^2}.$$
which was to be shown.

3rd solution. We have $\angle BAD + \angle DEB = \frac{\pi}{2} + \frac{\pi}{2} = \pi$, so $ABED$ is a cyclic quadrilateral. By Ptolemy's theorem we have $|AB| \cdot |DE| + |AD| \cdot |BE| = |AE| \cdot |BD|$ or
$$\frac{|AE|}{|BE|} = \frac{|AB| \cdot |DE|}{|BD| \cdot |BE|} + \frac{|AD|}{|BD|}.$$
Let $T$ be the intersection of the diagonals. Then $ABD$ and $TBA$ are right triangles and $\angle TBA = \angle ABD$, so they are similar. This implies that $\frac{|AD|}{|BD|} = \frac{|TA|}{|AB|}$ and $\frac{|AB|}{|BD|} = \frac{|BT|}{|AB|}$. Similarly, $DEB$ and $CTB$ are right triangles and since $\angle EBD = \angle TBC$, they too are similar. We see that $\frac{|DE|}{|BE|} = \frac{|CT|}{|BT|}$ and
$$\frac{|AE|}{|BE|} = \frac{|BT| \cdot |CT|}{|AB| \cdot |BT|} + \frac{|TA|}{|AB|} = \frac{|CT| + |TA|}{|AB|} = \frac{|CA|}{|AB|}$$
Finally, the triangles $ADC$ and $BAD$ are similar and we have $\frac{|CD|}{|AD|} = \frac{|AD|}{|AB|}$, so
$$\frac{|AE|}{|BE|} = \frac{|CA|}{|AB|} = \frac{|CA| \cdot |CD|}{|AD|^2} = \frac{|CA| \cdot |CD|}{|AC|^2 - |CD|^2}$$