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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.

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7 Answers 7

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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).

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    $\begingroup$ 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 $\angle DEA + \angle FEC + \angle AEF = 180^\circ$. $\endgroup$ Feb 9, 2021 at 14:44
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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: $$AE^2+EF^2=\frac{25}{16}a^2=AF^2$$

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

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

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

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  • $\begingroup$ I didn't see your answer at the first time, it's a smart one. $\endgroup$
    – PNT
    Feb 10, 2021 at 17:51
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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.

enter image description here

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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

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

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

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