Let $ABCD$ be a square and $E\in CD$ such that $DE=EC$. Let $F\in BC$ such that $BC=4FC$. Prove that $\angle AEF =90^\circ$.
My attempt:
Proving that $\angle AEF =90^\circ$ is the same as proving that $\triangle AEF$ is a right triangle. In other words, we wish to prove that $AE^2 +EF^2 = AF^2$. I’ve tried a lot of methods to reach this point but none of them worked.
Best Answer
Triangles EDA FCE are similar. Scale factor =2.
$$ FC= EC/2,\;DE= AD/2 \;$$
Since the triangles are similar corresponding angles are same. (Given square contains a right angle).