Balti Tee 1996 · Ülesanne 4

Solution:

The property that $\angle CPD$ is maximal is equivalent to the property that the circle $CPD$ touches the line $AB$ (at $P$). Let $O$ be the intersection point of the lines $AB$ and $CD$, and let $\ell$ be the bisector of $\angle AOD$. Let $A'$, $B'$ and $Q'$ be the points symmetrical to $A$, $B$ and $Q$, respectively, relative to the line $\ell$. Then the circle $AQB$ is symmetrical to the circle $A'Q'B'$ that touches the line $AB$ at $Q'$. We have

$$\frac{|OD|}{|OA'|} = \frac{|OD|}{|OA|} = \frac{|OC|}{|OB|} = \frac{|OC|}{|OB'|}$$

Hence the homothety with centre $O$ and coefficient $|OD|/|OA|$ takes $A'$ to $D$, $B'$ to $C$, and $Q'$ to a point $Q''$ such that the circle $CQ''D$ touches the line $AB$, and thus $Q''$ coincides with $P$. Therefore $\angle AQB = \angle A'Q'B' = \angle CQ''D = \angle CPD$ as required.