Angle chase:In $\Delta ABC, AB=AC $ and $\angle BAC=20°.$ If $CD$ is the median from $C$ to side $AB$, find $\angle ADC$. In $\triangle ABC, AB=AC $ and $\angle BAC=20^\circ$ If $CD$ is the median from $C$ to side $AB$, find $\angle ADC$.
 A: Notice, we have $$\angle A+\angle B+\angle C=180^\circ$$ But $AB=AC\iff \angle B=\angle C$ Hence, we get $$\angle A+\angle C+\angle C=180^\circ$$ $$20^\circ+2\angle C=180^\circ$$ $$\angle C=\frac{180^\circ-20^\circ}{2}=80^\circ$$
Let the unknown angle $\angle ADC=x$
Applying sine rule in $\triangle ACD$ as follows $$\frac{\sin\angle ACD}{AD}=\frac{\sin\angle CAD}{CD}$$ Setting the corresponding values, we get 
$$\frac{\sin(180^\circ-(x+20^\circ))}{AD}=\frac{\sin 20^\circ}{CD}$$ 
$$\frac{\sin(x+20^\circ)}{AD}=\frac{\sin 20^\circ}{CD}\tag 1$$ 
Similarly, applying sine rule in $\triangle BCD$ as follows $$\frac{\sin\angle BCD}{BD}=\frac{\sin\angle CBD}{CD}$$ Setting $BD=AD$ & the corresponding values, we get 
$$\frac{\sin(x-80^\circ)}{AD}=\frac{\sin 80^\circ}{CD}\tag 2$$ 
Now, diving (2) by (1) we get 
$$\frac{\frac{\sin(x-80^\circ)}{AD}}{\frac{\sin(x+20^\circ)}{AD}}=\frac{\frac{\sin 80^\circ}{CD}}{\frac{\sin 20^\circ}{CD}}$$
$$\frac{\sin(x-80^\circ)}{\sin(x+20^\circ)}=\frac{\sin 80^\circ}{\sin 20^\circ}$$
$$\frac{\sin x\cos 80^\circ-\cos x\sin 80^\circ}{\sin x\cos 20^\circ+\cos x\sin 20^\circ}=\frac{\cos 10^\circ}{\sin 20^\circ}$$
$$\frac{\tan x\sin 10^\circ-\cos 10^\circ}{\tan x\cos 20^\circ+\sin 20^\circ}=\frac{\cos 10^\circ}{\sin 20^\circ}$$ $$(\cos 20^\circ\cos 10^\circ-\sin 20^\circ\sin 10^\circ)\tan x=-2\sin 20^\circ\cos 10^\circ$$
$$\cos 30^\circ\tan x=-2\sin 20^\circ\cos 10^\circ$$
$$\tan x=\frac{-2\sin 20^\circ\cos 10^\circ}{\cos 30^\circ}=\frac{-2\sin 20^\circ\cos 10^\circ}{\frac{\sqrt 3}{2}}$$
$$\tan x=\frac{-4\sin 20^\circ\cos 10^\circ}{\sqrt 3}$$$$\iff x=\tan^{-1}\left(\frac{-4\sin 20^\circ\cos 10^\circ}{\sqrt 3}\right)$$
$$=180^\circ-\tan^{-1}\left(\frac{4\sin 20^\circ\cos 10^\circ}{\sqrt 3}\right)$$
Hence, we get $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\angle ADC}=\color{blue}{180^\circ-\tan^{-1}\left(\frac{4\sin 20^\circ\cos 10^\circ}{\sqrt 3}\right)\approx 142.12^\circ}}$$
A: ADC = t , ACD = x
AB = AC , AD = BD
In ACD : AC / sint = AD / sin(80-x)
In ABC : AC / sin 80 = BC / sin 20
In BDC : AD / sin 80 = BC / sin(180-x)
x = 60 , t = ADC = 140
