OSN 2007 Problem 1

Let $ABC$ be a triangle with $\angle ABC=\angle ACB=70^{\circ}$ . Let point $D$ on side $BC$ such that $AD$ is the altitude, point $E$ on side $AB$ such that $\angle ACE=10^{\circ}$ , and point $F$ is the intersection of $AD$ and $CE$ . Prove that $CF=BC$ .