### Problem Equilateral triangle $D E F$ is inscribed in equilateral triangle $A B C$ as shown with $E D \perp \overline{B C}$. What is the ratio of the area of $\triangle D E F$ to the area of $\triangle A B C ?$ ![[alc-eqT-insc-in-eqT.svg]] #### Solution No lengths are given, and a ratio is requested. The first length can be chosen. Let $m\overline{CD}=1$ Interior angles of an [[Equilateral Triangle]] like $\angle C$ are $60\degree$ Combined with $\angle EDC=90\degree$, $\triangle EDC$ is [[Special Right Triangles#30-60-90]] This means $m\overline{EC}=2$ and $m\overline{ED}=\sqrt3$ ![[alc-eqT-insc-in-eqT-1.svg]] The scale factor for the side lengths of the ([[‡ Similarity|similar]]) equilateral triangles $\triangle D E F$ to $\triangle A B C$ is $\frac{\sqrt{3}}{3}$, which, according to the [[Square-Cube law]], makes the ratio of the area of $\triangle D E F$ to the area of $\triangle A B C$ $\left(\frac{\sqrt{3}}{3}\right)^2=\boxed{\frac{1}{3}}$,