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.
Best Answer
Applying the trigonometric form of the Ceva's theorem we obtain:
In $\triangle ADB,\angle ADB=(180-18-30)^\circ=132^\circ$
Applying sine law in $\triangle ADB,$ $$\frac{AB}{\sin 132^\circ}=\frac{AD}{\sin30^\circ}\implies AD=\frac{AB}{2\sin48^\circ}$$ as $\sin132^\circ=\sin(180-132)^\circ=\sin48^\circ$
Best Answer
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: $$ \angle ACD=30^\circ,\quad \angle BCD=10^\circ. $$