Compute $m ( \angle ACD $).

Question

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 $.

Answer 1

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 $x=10^\circ$.

Answer 2

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

Answer 3

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 $\angle ACD = 40^{\circ}$.