, 2011

Let $T$ be the intersection of $BF$ and $CE$. We have that $\angle BCE + \angle BEC = \angle ABC$, and since $BC = BE$, it follows that $\angle BCE = \frac{1}{2} \angle ABC$. Similarly, $\angle CBF = \frac{1}{2} \angle ACB$. Hence
$$\angle CTF = \angle BCE + \angle CBF = \frac{1}{2}(\angle ABC + \angle ACB) = 60^\circ,$$
and so $ABTC$ is a cyclic quadrilateral. Hence $\angle EBF = \angle ACE = \angle AKE$ and $\angle ABF = 180^\circ - \angle EBF = 180^\circ - \angle AKE = \angle AKF$, which implies that $ABKF$ is cyclic. Hence we have that $\angle EBK = \angle CFK$, and since $ACKE$ is cyclic, we have $\angle BEK = \angle FCK$. Together with $BE = FC$ it follows that triangles $KBE$ and $KFC$ are congruent. Hence $KC = KE$, which means that $\angle CAK$ and $\angle BAK$ are subtended by equal chords in the circumcircle of $ACE$, and hence are equal, as required.

Let $L$ be the intersection of $EF$ and the angle bisector of $\angle BAC$. Let $BC = a$, $AC = b$ and $AB = c$. Then $EB = FC = a$.
Using the cosine rule in triangle $ABC$, we obtain $a^2 = b^2 + c^2 - bc$, and using the cosine rule in triangle $AEF$, we obtain
$$EF^2 = (a+c)^2 + (a+b)^2 - (a+c)(a+b) = a^2 + b^2 + c^2 + ab + ac - bc.$$
Since $AL$ bisects angle $AEF$, it follows that $\frac{EL}{FL} = \frac{AE}{FA} = \frac{a+c}{a+b}$, which yields $FL = \frac{a+b}{2a+b+c} FE$. Now,
$$\begin{align*} FL \cdot FE - FC \cdot FA &= \frac{a+b}{2a+b+c} FE \cdot FE - a(a+b) \\ &= \frac{a+b}{2a+b+c} (a^2 + b^2 + c^2 + ab + ac - bc) - a(a+b) \\ &= \frac{a+b}{2a+b+c} (b^2 + c^2 - bc - a^2) \\ &= 0 \quad (\text{cosine rule in triangle } ABC). \end{align*}$$
Hence $FL \cdot FE = FC \cdot FA$, which shows that $ACLE$ is a cyclic quadrilateral, i.e. $L = K$.