# Compute $m ( \angle ACD$).

Let $$\triangle ABC$$ s.t $$m (\angle A)=100^{°}, m (\angle B)=20^{°}$$. Let $$D\in Int (\triangle ABC)$$ s.t. $$m (\angle BAD)=30^{°}$$ and $$[BD$$ is the bisector of $$\angle B$$.

Compute $$m ( \angle ACD$$). I need a construction to solve it. I take $$M\in [AB]$$ s.t. $$m (\angle ACM)=40^{°}$$ and $$P\in [CM]$$ s.t. $$m (\angle MAP=30^{°}$$ I need to prove that $$[BP$$ is the bisector of $$\angle B$$. Then $$P=D$$.

We can use the Ceva's theorem also by the following way. Let $$\measuredangle ACD=x$$. So $$\measuredangle DCB=60^{\circ}-x$$ and we obtain: $$\sin{x}\sin10^{\circ}\sin30^{\circ}=\sin(60^{\circ}-x)\sin10^{\circ}\sin70^{\circ}$$ or $$\frac{1}{2\sin70^{\circ}}=\sin60^{\circ}\cot{x}-\frac{1}{2}$$ or $$\cot{x}=\frac{1+\sin70^{\circ}}{\sqrt3\sin70^{\circ}},$$ which gives $$x=40^{\circ}.$$

Indeed, $$\frac{1+\sin70^{\circ}}{\sqrt3\sin70^{\circ}}=\frac{1+\cos20^{\circ}}{2\cos30^{\circ}\cos20^{\circ}}=\frac{2\cos50^{\circ}+\cos30^{\circ}+\cos70^{\circ}}{4\cos50^{\circ}\cos30^{\circ}\cos20^{\circ}}=$$ $$=\frac{\cos50^{\circ}+\cos30^{\circ}+\cos10^{\circ}}{4\sin40^{\circ}\cos30^{\circ}\cos20^{\circ}}=\frac{2\cos30^{\circ}\cos20^{\circ}+\cos30^{\circ}}{4\sin40^{\circ}\cos30^{\circ}\cos20^{\circ}}=$$ $$=\frac{2\cos20^{\circ}+1}{4\sin40^{\circ}\cos20^{\circ}}=\frac{\cos20^{\circ}+\cos60^{\circ}}{2\sin40^{\circ}\cos20^{\circ}}=\frac{2\cos40^{\circ}\cos20^{\circ}}{2\sin40^{\circ}\cos20^{\circ}}=\cot40^{\circ}.$$

Let $$D'$$ be such that $$\angle BAD' = 30^{\circ}$$ and $$\angle BCD' = 20^{\circ}$$ and let us prove $$D'=D$$.

We see that now $$\angle ACD' = 40^{\circ}$$ so $$\angle CD'A = 70^{\circ}$$ and thus $$CA = CD'$$.

Rotate $$C$$ around $$D'$$ for $$60^{\circ}$$ to a new point $$E$$.

Then $$\angle ACE = 20^{\circ}$$ and $$CE = CD' = CA$$ so $$\angle AEC = \angle EAC = 80^{\circ}$$ and thus $$E,A,B$$ are collinear.

Now we see also that $$\angle ECB = 80^{\circ}$$ and $$\angle EAB = 20^{\circ}$$ so $$\angle BEC = 80^{\circ}$$ which means $$BE = BC$$.

So triangles $$EBD'$$ and $$CBD'$$ are congruent (sss), so $$BD'$$ is angle bisector of angle $$\angle ABC$$ which means $$D'=D$$.

So the angle we are looking for is $$\angle ACD = 40^{\circ}$$.

You can use the trigonometric form of Ceva's theorem (in your second setting): $$\frac{\sin\angle PAB}{\sin\angle PBA}\cdot\frac{\sin\angle PBC}{\sin\angle PCB}\cdot\frac{\sin\angle PCA}{\sin\angle PAC}=1,$$ i.e. $$\frac{\sin 30^\circ}{\sin x}\cdot\frac{\sin (20^\circ-x)}{\sin 20^\circ}\cdot\frac{\sin 40^\circ}{\sin 70^\circ}=1,$$ where $$x=\angle PBA$$. By writing $$\sin 30^\circ=1/2$$, $$\sin 70^\circ=\cos 20^\circ$$ and $$\sin 40^\circ=2\sin 20^\circ\cos 20^\circ$$, we get $$\sin(20^\circ-x)=\sin x$$ which has solutions $$x=10^\circ+k\pi$$, and since $$x$$ is an angle we have $$x=10^\circ$$.

• @MarianD It is not. Read it carefully. – SMM Mar 4 at 21:09
• @MarianD Exactly, my solution shows that $P$ and $D$ coinsides. The definitin of $P$ doesn't use the fact that it is on the bisector of $\angle B$. – SMM Mar 4 at 21:18
• You are right, excuse me (+1). – MarianD Mar 4 at 21:18