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Given a square $ABCD$, there is a point $E$ such that $\angle EDA = \angle ECB = 15^\circ$, find $\angle AEB$.

I placed the square with $CD$ being on the x axis and point $E$ on the positive y axis. That way I can find the position of $E$ using some simple trigonometry in a right triangle, and then calculate the distance from $E$ to either $A$ or $B$, which is exactly $1$. if $EA$ is the hypotenuse of a right triangle with the leg parallel to the x axis being $\frac12$, the triangle is a 30-60-60.

So half $\angle AEB=30^\circ$ and $\angle AEB=60^\circ$

But is there a way to find it without trigonometry? We didn't learn it in class yet, so there must be a way.

aradarbel10
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    Hint. Consider a point $F$ inside the square such that $DF=FC=a$, where $a$ is the length of the square edge, and prove that $DAEF$ is a rombus. – SMM Dec 19 '19 at 13:54
  • Dissect a regular dedecagon by cutting out six squares built on alternating sides. The remaining pieces are six equilateral triangles built on the remaining sides and a central regular hexagon. $ABCD$ is one of the squares. $E$ lies on two angle bisectors of the dodecagon and thus is at the center of both the dedecagon and the above-mentioned hexagon. – Oscar Lanzi Dec 19 '19 at 20:54

2 Answers2

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Connect E to midpoint F of AB;

$\angle FEC=\angle ECB=15^o$

Draw a circle on midpoint H of EB; clearly it passes point F and we get:

$\angle FHG=2 \angle FEH = 30^o $

$\angle FHG = \angle GHB = 30$

because HG is parallel with EF. Therefore we have:

$\angle FEB =\frac{1}{2} (2\times 30)=30^o$

The sketch is symmetric about EF , that is :

$\angle AEB= 2\times \angle FEB=60^o$

sirous
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Let $X$ be the point in the half plane determined by $AB$ which does not contain $C,D$, so that $\Delta AXB$ is equilateral.

A square and an equilateral triangle on top of it

The two triangles $\Delta ADX$ and $\Delta BCX$ are isosceles with an angle of $150^\circ$ in $A$, respectively $B$. This implies that the four acute angles in these two triangles have each $(180^\circ-150^\circ)/2=15^\circ$.

So the point $X$ is the point $E$ from the OP, and the angle $\widehat{AEB}=\widehat{AXB}$ is $60^\circ$ (by the construction of $X$).

dan_fulea
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