Find the angle in the drawing 
In the below drawing, we are given:
Triangle ABC is right. D is the midpoint of AB. Angle $ACD = φ$ and $BCE = φ$.
Angle $EAB = φ$.
$ED=α$ and $CD=4a$.
We are looking for angle φ (to be solved by using Geometry, not trigonometry).
What I have tried so far:
Since angles EAB and ECB are both φ, and AB is vertical to BC, then their other sides must also be vertical, so AEC is right.
Therefore the quadrilateral AEBC is inscribed in a semicircle (which I have drawn to see if I get any clue out of it) and AC is its diameter.
If G is the center of the circle, then EG and BG are equal to the circle radius.
Also angle $BDC = φ + BAC = EAC$ so triangles EAC and BDC are similar.
But I don't know how to use $ED=a$ and how to combine all this, to calculate angle φ.
By the way, I have found it in Geogebra to be ~19 degrees.
 A: Alright, I constructed the geometry and verified that the angle is 19 degrees, and I can't tell you how to calculate this angle without trigonometry, but I can tell you how this structure is built, and what the lengths $ \alpha $ and $ 4 \alpha $ reveal. Buckle up!
$$ \widehat {ACE} = \widehat {DCB} $$
Therefore the right triangles $ ACE $ and $ DCB $ are similar. As a result:
$$ \frac{AC}{DC} = \frac{AE}{DB} \qquad(1)$$
Now recall that $ CD $ is the median of $ AB $, so
$$ AD = DB $$
And replacing in (1) we will have:
$$ \frac{AC}{DC} = \frac{AE}{AD} \qquad(2)$$
Now consider the triangles $ACD$ and $EAD$. They have an equal angle ($\varphi$) and equal proportions of the lengths of the sides of that angle, as shown in (2). Therefore these two triangles are similar. And here is an interesting result:
$$ \frac{DC}{AD} = \frac{AD}{DE} $$
Or
$$ AD^2 = DC.DE $$
Now, recalling that $DE = \alpha$ and $DC = 4\alpha = 4DE$  we have:
$$ AD^2 = 4DE^2 $$
$$ AD = 2DE \qquad(3) $$
Et Voila!
$$ DC = 4DE = 2AD = 2DB \qquad(4) $$
Look at the right triangle $DBC$ : side $DB$ is half the hypothenuse $CD$ , which means angle $\widehat{DCB}$ must be 30 degrees.
So, to construct the angle $\varphi$ , all we need to do is draw the right triangle $DBC$ , extend the side $BD$ by equal length from point $D$ to find point $A$ , and connect $A$ to $C$ . The angle $\widehat{ACD}$ will be our $\varphi$ .
