Japan Mathematical Olympiad

$\frac{14\sqrt{65}}{13}$

By the alternate segment theorem and the assumption $AB=AC$, we have $\angle EFB=\angle EBA=\angle ACB$. Therefore, the four points $A$, $B$, $F$, $C$ are concyclic.

Let $X$ be any point on the tangent to $\omega$ at $E$, lying on the same side of line $AF$ as $B$. Then, by the alternate segment theorem, we have $$\angle DEB=\angle DEX+\angle XEB=\angle DAE+\angle EFB=\angle DAE+\angle EBA=\angle CEA.$$ Since $\angle DBE=\angle ACE$, we conclude that triangles $DEB$ and $AEC$ are similar. Therefore, we have $EB:EC=DB:AC=2:5$. Hence, we may write $EB=2x$ and $EC=5x$ with some $x>0$.

Since $A$, $B$, $F$, $C$ are concyclic, triangles $EBA$ and $EFC$ are similar. Therefore, we have $$EB:EF=EA:EC=AB:CF=5:10=1:2,$$ and hence, we have $EF=4x$ and $EA=\frac{5}{2}x$. By the power of a point theorem, we have $$25=A{B}^2=AE\cdot AF=\frac{5}{2}\cdot \frac{13}{2}x.$$

Solving this equation with the condition $x>0$, we get $x=\frac{2\sqrt{65}}{13}$. Thus, we conclude that $$ BC = BE + EC = 7x = \frac{14\sqrt{65}}{13}.