NZMO Round Two

Solution:
Let $x = \angle ABF$ and let $y = \angle BCA$. Since $\triangle ABF$ is isosceles, we get $\angle BFA = x$. Since $AB = AE$, it follows that arcs $BA$ and $AE$ are equal. Since equal arcs subtend equal angles, every angle subtended by either arc $AB$ or $AE$ must be equal to $\angle BCA = y$.

$$\Rightarrow \angle BCA = \angle BDA = \angle ACE = \angle ADE = y.$$

Since opposite angles in a cyclic quadrilateral ($ABDE$) are supplementary, we get $\angle ABD + \angle DEA = 180^\circ$. Therefore $\angle DEA = 180^\circ - x$. Now consider triangles $AED$ and $AFD$. We have

$$\angle ADE = y = \angle ADF \text{ and } \angle AED = 180^\circ - x = \angle AFD$$
and side $AD$ is shared. Therefore these triangles are congruent: $\triangle AED \equiv \triangle AFD$. Hence

$$\angle EAD = \angle DAF = x - y. \qquad (\text{angle sum in } \triangle AFD)$$

Now let $P$ be the intersection of $AD$ and $EF$. Also let $Q$ be the intersection of $AF$ and $CD$. Since $AP$ is the angle bisector of isosceles triangle $AEF$, we have

$$\angle APF = 90^\circ.$$
$$\Rightarrow \angle AFP = 90^\circ + y - x \qquad (\text{angle sum in } \triangle AFP)$$
$$\angle CFQ = 90^\circ + y - x \qquad (\text{vertically opposite})$$

Finally we get $\angle ECD = \angle EAD = x - y$ by the Bow Tie Theorem ($ACDE$ cyclic). Therefore $\angle FCQ = x - y$. Now consider the sum of the angles in triangle $FCQ$ to get

$$\angle CQF + (x - y) + (90^\circ + y - x) = 180^\circ$$

$$\Rightarrow \angle CQF = 90^\circ$$

as required.