1976 AHSME Problems/Problem 16

Let $AC=BC=DF=EF=x$ and let the altitude from $F$ to $\overline{DE}$ have length $y$. This gives that $AB=2y$. Since $\triangle ABC$ and $\triangle DEF$ are isosceles, we have that $\angle ACB = 2\sin ^{-1} \left(\frac{y}{x} \right)$ and $\angle DFE = 2\cos ^{-1} \left(\frac{y}{x} \right)$. Since the sum of the inverse sine and cosine of any possible sine/cosine fraction is $90^\circ$, we have that $m \angle ACB + m \angle DFE = 180^\circ$. Thus, $\angle ACB$ and $\angle DFE$ are supplementary, so $\mathbf{II.}$ is true while $\mathbf{I.}$ is not.

Applying the Pythagorean Theorem gives that the length of the altitude from $C$ to $\overline{AB}$ is $\sqrt{x^2-y^2}$ and $DE = 2 \sqrt{x^2 - y^2}$. This means that $[ABC]=[DEF]=y \sqrt{x^2-y^2}.$ So we also have that $\mathbf{III.}$ is true while $\mathbf{IV.}$ is not.

Thus, our answer is $\textbf{(E) }\textbf{II. }\text{and }\textbf{III. }\text{only.}$