1959 AHSME Problems/Problem 10

In $\triangle ABC$ with $\overline{AB}=\overline{AC}=3.6$, a point $D$ is taken on $AB$ at a distance $1.2$ from $A$. Point $D$ is joined to $E$ in the prolongation of $AC$ so that $\triangle AED$ is equal in area to $ABC$. Then $\overline{AE}$ is: $\textbf{(A)}\ 4.8 \qquad\textbf{(B)}\ 5.4\qquad\textbf{(C)}\ 7.2\qquad\textbf{(D)}\ 10.8\qquad\textbf{(E)}\ 12.6$


Note that $\frac{1}{2}AB * AC *\sin\angle BAC = \frac{1}{2}AD * AE *\sin\angle DAE$. Since $\angle BAC = \angle DAE$, we have $AB*AC = AD*AE$, so that $3.6*3.6 = 1.2*AE$. Therefore, $AE = \frac{3.6^2}{1.2} = 10.8$. Thusly, our answer is $\boxed{\text{(D)}}$, and we are done.