In quadrilateral $ABCD$, $\angle BAC=\angle CAD=2\,\angle ACD=40^\circ$ and $\angle ACB=70^\circ$. Find $\angle ADB$. 
Let quadrilateral $ABCD$ satisfy $\angle BAC = \angle CAD = 2\,\angle ACD = 40^\circ$ and $\angle ACB = 70^\circ$. Find $\angle ADB$.

What I tried

*

*Ceva’s Theorem (Trigonometry version)

*Try to construct some equilateral triangle.

Which both failed.
Any hints or solutions please. Thanks in advance!


Since $\angle ACB=\angle ABC=70^\circ$, the triangle $ABC$ is isosceles and $\;\overline{AB}=\overline{AC}$.
By applying the law of sines to the triangle $ACD$, we get that:
$\overline{AD}=\overline{AC}\cdot\cfrac{\sin\angle ACD}{\sin\angle ADC}=\overline{AC}\cdot\cfrac{\sin 20^\circ}{\sin 120^\circ}=\cfrac{2\overline{AC}\sin 20^\circ}{\sqrt{3}}\;.$
And, by applying the law of sines to the triangle $ABD$, we get that:
$\overline{AD}\sin\angle ADB=\overline{AB}\sin\angle ABD\;.\quad\color{blue}{(*)}$
Let $\;\alpha=\angle ADB\;.$
Since $\;\overline{AD}=\cfrac{2\overline{AC}\sin 20^\circ}{\sqrt{3}}\;$, $\;\overline{AB}=\overline{AC}\;$ and $\;\angle ABD=100^\circ-\alpha\;,\;$ the equality $(*)$ turns into:
$\cfrac{2\overline{AC}\sin 20^\circ\sin\alpha}{\sqrt{3}}=\overline{AC}\sin(100^\circ-\alpha)\;,$
$2\sin 20^\circ\sin\alpha=\sqrt{3}\sin(90^\circ+10^\circ-\alpha)\;,$
$4\sin 10^\circ\cos 10^\circ\sin\alpha=\sqrt{3}\cos(10^\circ-\alpha)\;,$
$4\sin 10^\circ\cos 10^\circ\sin\alpha=\sqrt{3}\left(\cos10^\circ\cos\alpha+\sin 10^\circ\sin\alpha\right)\;,$
$4\sin 10^\circ\sin\alpha=\sqrt{3}\left(\cos\alpha+\tan 10^\circ\sin\alpha\right)\;,$
$\left(4\sin 10^\circ-\sqrt{3}\tan 10^\circ\right)\sin\alpha=\sqrt{3}\cos\alpha\;,$
$\tan\alpha=\cfrac{\sqrt{3}}{4\sin 10^\circ-\sqrt{3}\tan 10^\circ}\;.$
Hence,
$\angle ADB=\alpha=\arctan\left(\cfrac{\sqrt{3}}{4\sin 10^\circ-\sqrt{3}\tan 10^\circ}\right)\simeq\\\simeq 77,3361794^\circ.$
