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.

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).

• Just to make it crystal-clear: since the triangles are the similar and since they're right triangles, $90^\circ = \angle DAE + \angle DEA = \angle FEC + \angle DEA$. And $\angle DEA + \angle FEC + \angle AEF = 180^\circ$. Feb 9 '21 at 14:44

The idea in your attempt definitely works here (although it is not the shortest proof, see the answer by Narasimham): Let $$a$$ be the sidelength of the square. Then we have \begin{align*} AE^2&=AD^2+DE^2=a^2+\left(\frac{1}{2}a\right)^2\\ EF^2&=EC^2+CF^2=\left(\frac{1}{2}a\right)^2+\left(\frac{1}{4}a\right)^2\\ AF^2&=AB^2+BF^2=a^2+\left(\frac{3}{4}a\right)^2 \end{align*} And thus: $$AE^2+EF^2=\frac{25}{16}a^2=AF^2$$

$$\tan(\angle AED) = 2.$$
$$\tan(\angle FEC) = (1/2) =$$cot$$(\angle AED).$$

Further, $$\angle AED$$ and $$\angle FEC$$ are both acute.

Therefore, $$\angle AED + \angle FEC = 90^\circ.$$

• I didn't see your answer at the first time, it's a smart one. Feb 10 '21 at 17:51

To follow the OP first intuition.

Consider $$AB=4$$. So you have $$BF=3$$ and $$AF=5$$.

Apply Pythagoras' theorem to the two smaller triangles and verify that $$AEF$$ is a right triangle because it follows Pythagoras' theorem. Ignoring the similar triangles for variety...

Let the side length of the square be $$s$$. We have $$|AF|^2 = |AB|^2 +|BF|^2 = s^2 + \left(\frac 34 s\right)^2 = \frac{25}{16}s^2$$

Then $$|AE|^2 = |AD|^2 +|DE|^2 = s^2 + \left(\frac 12 s\right)^2 = \frac{5}{4}s^2 = \frac{20}{16}s^2$$
and $$|EF|^2 = |EB|^2 +|BF|^2 = \left(\frac 12 s\right)^2 + \left(\frac 14 s\right)^2 = \frac{5}{16}s^2$$

giving $$|AF|^2 = |AE|^2 +|EF|^2$$ as required

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 $$AEF = 90$$.

\eqalign{ & AE^2 + EF^2 = \cr & AD^2 +DE^2 + EC^2 + CF^2 = \cr & a^2 + {a^2 \over 4} + {a^2 \over 4} + {a^2 \over 16} = {25a^2 \over 16} \cr & \cr & AF^2 = AB^2 + BF^2 = a^2 + {9a^2 \over 16} = {25a^2 \over 16} \cr }

since $$AE^2 + EF^2 = AF^2 \rightarrow AE \perp EF$$