Find the missing angle in the triangle Given the following triangle $ABC$, find the angle of $ACD$.

Notice that $AC = BC$, I can find the orthocenter of $ABC$. However, I am stucked and do not know what is the next step. Any hint would be appreciated.

Applying the trigonometric form of the Ceva's theorem we obtain:
$$\frac{\sin\angle ACD}{\sin\angle BCD}=\frac{\sin40^\circ\sin50^\circ}{\sin30^\circ\sin20^\circ}=\frac{2\sin40^\circ\cos40^\circ}{\sin20^\circ}
=\frac{\sin80^\circ}{\sin20^\circ}=\frac{\cos10^\circ}{2\sin10^\circ\cos10^\circ}=\frac{\sin30^\circ}{\sin10^\circ}.$$
From this and $$\angle ACD+\angle BCD=40^\circ$$ one concludes:
$$
\angle ACD=30^\circ,\quad \angle BCD=10^\circ.
$$
