Geometry question about a triangle in a circle Circumscribe triangle $ABC$. Extend $B$ to a point $P$ on the circumference of the circle. Now extend the line $AB$ to a point $P_1$. Why is $\angle BAC + \angle BPP_1 + \angle BPP_3= 180^\circ$? ![](

Assuming that $P$ is arbitrary, and that $P_1$ and $P_2$ are then constructed so that $\angle AP_1P$ and $\angle AP_3P$ are right, we have that

$$\angle AP_1P + \angle P_1PP_3 + \angle PP_3A + \angle P_3AP_1 = 360^\circ$$ $$\angle P_1PP_3 + \angle P_3AP_1 = 180^\circ$$ $$(\angle BPP_1 + \angle BPP_3) + \angle CAB = 180^\circ.$$