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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 $\angle BAC + \angle BPP_1 + \angle BPP_3= 180^\circ$?

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Isn't B already on the circle? – user7530 Feb 15 '13 at 23:25
Yes, B is on the circle. I said circumscribe triangle ABC. The points A, B and C touch the circumference of the circle. – janice Feb 15 '13 at 23:26
Can you put some image with all elements and points on it? Mabe it's just me, but it's not really clear what exactly you're doing. – Kaster Feb 15 '13 at 23:27
The point P creates a cyclic quadrilateral ABPC. Hold on I will try to post a pic. – janice Feb 15 '13 at 23:28
So wait, $P_1$ is chosen so that $\angle BP_1P$ is right? – user7530 Feb 15 '13 at 23:44
up vote 1 down vote accepted

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

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

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My question is how do you know that $\angle P_1PP_3 + \angle P_3 AP_1 = 180$? The point $P_1$ is outside of the circle. – janice Feb 16 '13 at 0:07
I proved it starting from the fact (line 1) that any quadrilateral's interior angles sum to 360 degrees. I then subtracted (line 2) the two angles that are known to be 90 degrees. – user7530 Feb 16 '13 at 0:08

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