8. Point $\mathrm{D}$ is drawn on side $B C$ of equilateral triangle $A B C$, and $A D$ is extended past $D$ to $E$ such that angles $E A C$ and $E B C$ are equal. If $B E=5$ and $C E=12$, determine the length of $A E$.
Solution:By construction, $A B E C$ is a cyclic quadrilateral. Ptolemy's theorem says that for cyclic quadrilaterals, the sum of the products of the lengths of the opposite sides equals the product of the lengths of the diagonals. This yields $(B C)(A E)=(B A)(C E)+(B E)(A C)$. Since $A B C$ is equilateral, $B C=A C=A B$, so dividing out by this common value we get $A E=C E+B E=17$.