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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.

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2 Answers 2

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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.

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  • $\begingroup$ It could be made clearer that $BAE''$ being isosceles follows from the fact that it has angles $\angle BAE''=90-60=30$ and $\angle ABE''=90-15=75$. $\endgroup$ Jun 29 at 7:47
  • $\begingroup$ Ok, edited. Thanks, @JaapScherphuis. $\endgroup$
    – loopy walt
    Jun 29 at 8:01
  • $\begingroup$ @loopywalt I don't understand the final part of the argument. What do you mean that $E'$ and $E''$ can only be identical if they coincide with $E$?. This is where I am confused. Can you please attach a diagram for this?. I mean to follow your steps I drawn it but because it appears three points $E$, $E'$ and $E''$. This makes it confusing. Because of this I am requesting a separate drawing with the second step, I mean the part where you establish that $E=E'$. Doing this will help me to understand better. $\endgroup$ Jul 1 at 22:02
  • $\begingroup$ @loopywalt What is the interpretation of the intersection?. Do you mean the extension of the lines or what?. This is confusing. $BE$ could be larger or shorter than $BE'$. If $BE$ is shorter then it may be the extension of BE if is larger then it is just that point. In one way or another this intersection becomes $E''$. That's how I am understanding it. Therefore $E'=E''$. $\endgroup$ Jul 1 at 22:07
  • $\begingroup$ @loopywalt Then $AB=AD$ is because $ABCD$ is a square. $AE''=AB$ because $\triangle{ABE''}$ is isosceles. Then $E'$, why are you bringing up this?, we already established that $E'=E''$. You say $E'$ belongs to $DE$, because we already established this first. And $E''$ belongs to $BE$ but what? The extension or what?. Then there is the final argument which says that because they can only be identical if they coincide with $E$. $\endgroup$ Jul 1 at 23:39
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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:


enter image description here

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 $\angle{CBE}=15^{\circ}$.

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