I already have my own solution for the following question. But I am still interested in other elegant solutions without trigonometry if possible.
This is my own solution. I am lazy to upload the TeX code, I am sorry.
I already have my own solution for the following question. But I am still interested in other elegant solutions without trigonometry if possible.
This is my own solution. I am lazy to upload the TeX code, I am sorry.
A hint:
Draw a regular $18$-gon $Q$. It has the property that the angle between neigbouring diagonals emanating from the same vertex is $10^\circ$.
The points $B$, $C$, $D$ of your figure can be realized as vertices of $Q$; furthermore the line $C\vee P$ is a diagonal emanating from $C$, and $D\vee A$, $\ D\vee E$ are diagonals emanating from $D$.
I think that the solution of your problem is hidden in this figure. The nontrivial point is the fact that the line $B\vee P$ is also a diagonal, i.e, that three diagonals of $Q$ meet at $P$. This in turn has to do with algebraic relations among the $18$th roots of unity.
Let's try for $\alpha$ ...
In $\triangle BPQ$, we have $\angle B = \alpha$ and $\angle P = 120 - \alpha$.
$\begin{eqnarray*} \frac{\sin(120-\alpha)}{\sin\alpha}=\frac{BQ}{PQ}=\frac{BQ}{CQ} \frac{CQ}{DQ} \frac{DQ}{PQ}=\frac{\sin 40}{\sin 20}\frac{\sin 50}{\sin 70} \cdot 1=\frac{2\sin 40 \cos 40}{2\sin 20 \cos 20}=\frac{\sin 80}{\sin 40} \end{eqnarray*}$
Observe that $80 + 40 = 120$. Thus,
$\begin{eqnarray*} \sin(120-\alpha) \sin 40 &=& \sin \alpha \sin( 120-40 ) \\ (\sin 120 \cos\alpha - \cos 120 \sin \alpha ) \sin 40 &=& \sin\alpha ( \sin 120 \cos 40 - \cos 120 \sin 40 ) \\ \cos\alpha \sin 40 &=& \sin\alpha \cos 40 \\ 0&=& \sin( \alpha - 40 ) \\ \alpha &=& 40 \text{ is the only possible answer} \end{eqnarray*}$
Note: Generalizing $120$ to an angle $\gamma$ such that $\sin{\gamma} \neq 0$, we have
$\frac{\sin(\gamma-\alpha)}{\sin\alpha} = \frac{\sin(\gamma - \beta)}{\sin\beta} \implies \sin(\alpha-\beta) = 0$
Second part.
Construct equilateral triangle DEF. F is point on EB.
Rotate triangle EDF about point E clockwise 20 deg so that F falls at F' and D falls on DB at D'.
Since angle DEF' = 80 deg, then angle EDF' = 50 deg.
Because angle EDA = 50 deg.
Therefore D, A and F' are colinear. ......(1)
angle BED = angle EBD = 40 deg
So D'B = D'E = D'F'
Since angle BD'F' = 40
Therefore angle D'BF' = 70 i.e. angle EBF' = 30 deg
angle EBA = 30 deg (given)
Therefore B, F' and A are colinear. ......(1)
From (1) and (2) we know that A and F' are the same point.
Therefore angle EAD = 50 deg
Construct equilateral triangle PDF (where F is below BC)
Join FB and FC
Triangles FPC and DPC are congruent (SAS)
Therefore CF = CD
angle CFD = angle CDF = 20 deg.
Then C,F,B, D concylic.
angle FBC = angle FDC = 20 deg
Triangle BPD and BFD are congruent (SAS)
Therefore angle PBD = 40 deg.