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

Well, Geogebra says it is $\approx 77,34^{\circ}$, so good luck...

Actually, Ceva might really help:

$${\sin 80\over \sin 40}{\sin(70-x)\over \sin x}{\sin 20\over \sin90} = 1$$

After some manipulation we get $$\cot x = \tan 20+{2\over \cos 10}\implies x =... $$

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,

$\angle ADB=\alpha=\arctan\left(\cfrac{\sqrt{3}}{4\sin 10^\circ-\sqrt{3}\tan 10^\circ}\right)\simeq\\\simeq 77,3361794^\circ.$