Need a pure geometric solution to a $20-30-130$ triangle question

Question

In $\triangle{ABC}$, $\angle{ABC}=20^{\circ}$, $\angle{ACB}=30^{\circ}$, $D$ is a point inside the triangle and $\angle{ABD}=10^{\circ}$, $\angle{ACD}=10^{\circ}$, find $\angle{CAD}$.

Note: I have seen some very similar question with beautiful solution in pure geometric format. I know how to solve this problem in trigonometric format. But I think this problem deserves a beautiful geometric approach as solution, and that's why I post it here.

As request, here is approach applying Ceva's theorem in trigonometric form,

$$\begin{align*} \frac{\sin130}{\cos130+2\cos10}&=\tan(x)\Longrightarrow \frac{\sin120\cos10+\cos120\sin10}{\cos120\cos10-\sin120\sin10+2\cos10}\\ &=\frac{\sqrt{3}\cos10-\sin10}{3\cos10-\sqrt{3}\sin10}.\\ &=\frac{1}{\sqrt{3}}\\ &=\tan30\Longrightarrow x\\ &=\boxed{30}\\ \end{align*}$$

Answer 1

Please extend line segment $BA$. We have $\angle CAE = 50^0$. Draw $\angle ACE = 50^0$. We have $CE = AE$.

So, $\angle BCE = \angle BEC = 80^0$. $BM$ is angle bisector of isosceles triangle $\triangle CBE$ where $BC=BE$.

Therefore $CD = DE$. As $\angle DCE = 60^0$, $\triangle DCE$ is equilateral triangle and $DE = CE = AE$. So $\triangle AED$ is isosceles triangle with $\angle AED = 20^0$.

That leads to $\angle DAE = 80^0 \implies \angle DAC = 30^0$.

Answer 2

COMMENT.-This could be another nice way to prove that $x=30º$.