Points $E$ and $F$ are located on square $ABCD$ so that $\triangle BEF$ is equilateral. What is the ratio of the area of $\triangle DEF$ to that of $\triangle ABE$? $\mathrm{(A) \ } \frac{4}{3} \qquad$ $ \mathrm{(B) \ } \frac{3}{2} \qquad$ $ \mathrm{(C) \ } \sqrt{3} \qquad$ $ \mathrm{(D) \ } 2 \qquad$ $ \mathrm{(E) \ } 1+\sqrt{3}$
Solution
Let $AB=1$ and $AE=a$. Since $BE=BF$, right triangles $ABE$ and $CBF$ are congruent, so $CF=a$ and $DE=DF=1-a$.
$BE=EF$ gives $1+a^2=2(1-a)^2$, i.e. $a^2-4a+1=0$, so $a=2-\sqrt3$ and $DE=DF=\sqrt3-1$.
The ratio is $\frac12DE^2:\frac12AB\cdot AE=(\sqrt3-1)^2:(2-\sqrt3)=(4-2\sqrt3):(2-\sqrt3)=2$.
Rectangle $ABCD$ has $AB=8$ and $BC=6$. Points $E$ and $F$ lie on $\overline{AD}$ and $\overline{DC}$, respectively, and $\triangle BFE$ is an isosceles triangle with $\angle BFE=90^\circ$, as shown. What is the ratio of the area of $\triangle DEF$ to the area of $\triangle ABE$?
$\mathrm{(A) \ } \frac{4}{3} \qquad$ $ \mathrm{(B) \ } \frac{3}{2} \qquad$ $ \mathrm{(C) \ } \sqrt{3} \qquad$ $ \mathrm{(D) \ } 2 \qquad$ $ \mathrm{(E) \ } 1+\sqrt{3}$ 
