<s>[INST] Prove that $\angle AEF =90^\circ$ given a square $ABCD$ 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.
 [/INST] Let $AB = BC = CD = DA = 4x$.
Then: $DE = EC = 2x$, $CF = x$, and $FB = 3x$.
Now,

*

*In triangle $ABF$, we have $AF^2 = AB^2 + FB^2 = (4x)^2 + (3x)^2 \Rightarrow AF = 5x$.

*In triangle $ADE$, we have $AE^2 = DA^2 + ED^2 = (4x)^2 + (2x)^2 \Rightarrow AE = 2\sqrt{5}x$.

*In triangle $ECF$, we have $EF^2 = EC^2 + CF^2 = (2x)^2 + x^2 \Rightarrow EF = \sqrt{5}x$.

Now, if you look at triangle $AEF$, we have $\underbrace{AE^2}_{(2\sqrt{5}x)^2} + \underbrace{EX^2}_{(\sqrt{5}x)^2} = \underbrace{AF^2}_{(5x)^2}$. Since Pythagoras identity holds here, we can deduce that angle $AEF = 90$.
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