Triangle $ABC$ is inscribed in circle $\omega$. Point $P$ lies on line $BC$ such that line $\overline{PA}$ is tangent to $\omega$. The bisector of $\angle APB$ meets segments $\overline{AB}$ and $\overline{AC}$ at $D$ and $E$ respectively. Segments $\overline{BE}$ and $\overline{CD}$ meet at $Q$. Given that line $\overline{PQ}$ passes through the center of $\omega$, compute $\angle BAC$.
I tried drawing a diagram, but this led to many problems. I thought of using trigonometry, which seemed like a good idea at first, but then crashed and failed. People are recommending a topic called barycentric coordinates. Can someone guide me through the solution and then point me towards a credible source? Thank you!
Produce BO to cut the circle at X. Join CX. From A, drop the perpendicular to PE cutting the circle at Y. Join YB, YO, YX and YC.
After making the necessary constructions, we have all same color coded angles are equal. In particular, $\phi = 0.5 \theta$; and $\angle BCX = 90^0$ ….. (0).
$\phi_2 = \phi_3$ implies $OY // XC$. …. (1)
Let OY cut BC at Z. Because OBYC is a kite, $\angle OZC = 90^0$ ..... (2)
Let T be the midpoint of XC (i.e. XT = TC). Then $\angle OTC = 90^0$ ….. (3)
All of the above make sure that OZTX is a //gm; and OZCT is a rectangle. Therefore, OYCX is rhombus. This makes the already isosceles $\triangle XOC$ equilateral. Hence $\theta = 60^0$.