Find the angles without using trigonometry 
In the figure, given that $JA=JB$, $\angle BAK=20^{\circ} $, $\angle KAJ=50^{\circ} $, $\angle KBA=40^{\circ} $, Find $\angle BJK$.
I succeed in solving it using trigonometry (quite tedious), but the answer turn out to be very neat and i believe there must be a very simple way to solve it using plane euclidean geometry. (pretty sure it exist but i tried many construction and still don’t work) I want to see the solution without trigonometry.
Any hints or solution is greatly appreciated. BTW, the angle i found it $10^{\circ} $ (neat right?)
 A: Hint:
Triangle is isosceles so:
$\angle CBA=\angle CAB=70^o\Rightarrow \angle ACB=40^o$
Draw the circumcircle of ABC and mark it's center as O.
Draw a diameter from A to meet the circle at D. Clearly AD bisect angle ACB. We have:
$\angle BCD=\frac {40}2=20^o$
Draw radius OE such that it intersect BD, clearly OE is perpendicular bisector of chord BD, that means it is parallel with BC, because angle CBD is opposite to diameter CD , so it is $90^o$. In this way E is mid point of arc BD and we have:
$OE||CB\Rightarrow \angle DOE=\angle BCE=20^o$
If the extension of CK meets the circle at E that would mean CE is bisector of angle BCD and so $\angle BCK= \frac{20}2=10$
Now we have to show the extension of CK meets the circle at E:
In triangle ABK:
$\angle AKB=120^o$
Connect A and C to E,Mark intersection of AB with CE as G.
$\angle ACE=30^0$
In triangle ACG we have:
$\angle AGC=180-[(\angle ACG=30) +(\angle CAG=70)]=80^o $
So in triangle ACK'( K' is the intersection of CE with AK) we have:
$\angle AK'G = 180-(20+80)=80^o$
Angle K'GA is exterior angle of triangle BGK' so we have:
$\angle BK'G=80-40=40^o\Rightarrow \angle BK'A=40+80=120^o=\angle BKA$
that is K is coincident on K' and as a result the extension of CK meets the circle at E.
A: The analytic approach is short and easy.
The distance from K to the midline appears to equal the distance to BJ.
Which means that JK bisects angle BJF. 
A: Here is a solution not using trigonometry:
Assume $H$ is the midpoint of $AB$. Let's extend $BK$ such that it intersects $JH$ at $L$. Then, it is obvious that $\angle LAB=40^{\circ}$, so $\angle KAL=20^{\circ}$, which means $KA$ is the angle bisector of $\angle BAL$. Hence, by the angle bisector theorem, we have: $$\frac{LK}{KB}=\frac{AL}{AB}.$$

Note that this theorem can be proved without using trigonometry; for example take a
look at this link.


On the other hand, consider the figure below, and observe that $\angle LAS=30^{\circ}$. Thus $LS=\frac{AL}{2}$.

Therefore, since $\triangle JLS$ and $\triangle JHA$ are similar, we have:
$$\frac{JL}{LS}=\frac{JA}{AH} \\ \implies \frac{JL}{JA}=\frac{AL}{AB} \implies \frac{JL}{JB}=\frac{AL}{AB}=\frac{LK}{KB},$$
which implies that $JK$ is the angle bisector of $\angle BJL$. So, $\angle BJK=10^{\circ}$.
We are done.
