What is the measure of $\angle{BAE}$ and the length of $BE$ and $ED$ in a square?

Question

The puzzle is as follows:

Suppose that you have a square whose sides measure 1 inch. Let each vertex be $A$, $B$, $C$ and $D$. Now, pick a point $E$ on the interior of this square so that $\angle{EDA}=60^{\circ}$ and $\angle{CBE}=15^{\circ}$. With this in mind and relying only in synthetic geometry, that is no trigonometry allowed. Find the lengths of $BE$, $ED$ and the measure of $\angle{BAE}$.

This problem arised from a discussion and I am not sure how to prove the solution in a reasonable manner.

The original puzzle appears to be some adaptation of the works from George Birkhoff and Ralph Beatley's book of the 1950's and from Markus Horbit's Plane geometry outlines of 1984.

What I did was to "guess" that what if $ED=DA$. This would make $\triangle{AED}$ equilateral and with that $AB=AE=AD$ which make those being the circumradius of $\triangle{BCD}$.

This solves the problem, as $\angle{BAE}=30^{\circ}$. And as for the lenghts requested you may use ratios from special right triangles and their pythagorean triples.

I constructed a perpendicular from $E$ to $AD$ and called the intersection point $F$ and by the perpendicular bisector theorem, $FD=FA$.

I also constructed a perpendicular from $E$ to $BC$ with the intersection called $G$.

It turns out that $GE \parallel BC$, so $GEF$ are collinear which makes $FD=FA=BG=GC=\frac{1}{2}$.

This allows the computation of BE using pythagorean ratios.

In a 15-75-90 triangle these are, $\sqrt{6}-\sqrt{2}:\sqrt{6}+\sqrt{2}:4$.

From this it is known: $\frac{1}{2}=k(\sqrt{6}+\sqrt{2})$

Therefore: $4k=\frac{2}{\sqrt{6}+\sqrt{2}}$

And if you do use rationalization you could get:

$BE=\frac{\sqrt{6}-\sqrt{2}}{2}$ (also in inches)

But as you can see, the problem itself is how to prove that $ED=DA$.

Is there some sort of construction or line or circle here that can properly justify this?. Please keep in mind that I am not good with geometry so I appreciate somebody could attach a drawing or schematic so I could understand this better. Specially as I am requesting what sort of construction is needed.

Answer 1

Find E' inside the square such that DAE' is equilateral (this contains the line DE but it is not a priori clear that E=E'). Find the intersection E'' of AE' with EB. We'll show E=E'=E''.

E'=E'': BAE'' is isosceles (its angles are easily computed as 90°-60° = 30°, 90°-15° = 75° and 180°-30°-75° = 75°; clarification advocated by @JaapScherphuis), therefore the length AE''=AB=AD=AE' But E' lies on DE and E'' lies on BE. They can only be identical if they coincide with the intersection, i.e. E.

From this everything else follows easily.

Answer 2

I can't prove that $ED=DA=AE$, but can show that if it is, then the initial condition $\angle{EDA}=60^{\circ}$ and $\angle{CBE}=15^{\circ}$ is true, with this construction:

There are 6 equilateral triangles arranged around $A$ inside a circle.
There are also 12 isosceles triangles, forming a dodecagon.
Its interior angle is $ 2n - 4 $ right-angles, which is $150^{\circ}$.

So $\angle{ABE}=75^{\circ}$ and $\angle{CBE}=15^{\circ}$.