What-s the measure of the $\angle BAC$ in the triangle isosceles below?

Question

For reference: In the figure below $AB=AC$ and $P$ is the circumcenter of triangle $ABC$. It is also known that $AD=BC$. Calculate the measure of angle $BAC$ (Answer:$20^o$)

Progress I found

$\alpha = \frac{\angle BAC}{2}$

M is the midpoint of

$BC = 2x \Leftrightarrow BM = x\\ R=PA=PB=PC\\ \triangle PMC:R(sen(2\alpha)=x\\ \triangle APD: \frac{2x}{sen2\alpha}=\frac{R}{sen (180-3\alpha)}= \frac{R}{sen(3\alpha)}=\frac{x}{sen(2\alpha)sen(3\alpha}\implies\\ sen(3\alpha) = \frac{1}{2}=sen30^o \\ \therefore \angle BAC = 20^o$

Does anyone have a resolution by geometry?

Answer 1

Copy $\triangle BPC$, and add a point $O$ on the left side of $AD$ such that $\triangle DOA\cong \triangle BPC$.

Using the same notation as the question,

$$\angle DOA = \angle BPC = 4\alpha$$

As opposite angles at point $P$,

$$\angle DPA = \angle BPM = 2\alpha$$

As $OD=OA = R$, and with $\angle DOA = 2\angle DPA$, a circle centred at $O$ through $D$ and $A$ would pass through $P$, so

$$OP=OA=R = AP\\ \angle POA = 60^\circ$$

On the other hand, $\angle POD = 2\angle PAD = 2\alpha$, so

$$\begin{align*} \angle POD + \angle DOA &= \angle POA\\ 2\alpha + 4\alpha &= 60^\circ\\ \angle BAC = 2\alpha &= 20^\circ \end{align*}$$

Answer 2

This answer is based on the original proof in the question, which uses the sine laws. But are the sine laws not geometry? The sine laws may be proven using altitudes of the triangle, so:

Consider $\triangle APD$. Drop an altitude from $A$ onto the extended line $PD$, and let the foot be $N$.

$$\begin{align*}\triangle ANP &= 90^\circ\\ AP&= BP = R\\ \angle APN &= \angle BPM = 2\alpha\\ \triangle ANP &\cong \triangle BMP &&\text{(AAS)} \end{align*}$$

$AN=BM=x = \frac12 AD$, so $\triangle AND$ is a $60-90-30$ triangle.

As an external angle of $\triangle APD$,

$$\angle NDA = \alpha + 2\alpha = 30^\circ\\ \angle BAC = 2\alpha = 20^\circ$$