Circle of Apollonius proof question

Question

Im reading a proof of the Circle of Apollonius and I am unsure of one part of it -

Find the locus of a point P whose distances from two fixed points, A and A' are in a ratio of 1 : $\mu$.

Define a points A1 and A2 on line AA' such that A1P bisects the internal angle APA' and A2P bisects the external angle APA'. Then define points E and F on AP such that A'E is parallel to A1P and A'F is parallel to A2p, that is, perpendicular to A1P.

From this we have $\frac{AA1}{A1A'}=\frac{AP}{PE}=\frac{AP}{PA'}=\frac{AP}{FP}=\frac{AA2}{A'A2}=\frac{1}{\mu}$

Since $\angle A1PA2$ is a right angle, P lies on a circle with diameter A1A2.

My question is, how do we know $\angle A1PA2$ is a right angle? I can see how it follows that it is a right angle after saying A'F is perpendicular to A1p, but I dont understand how we get that either.

Edit: Sorry I don't have a diagram to go with this, its from my book. The diagram on the third page of this explanation is fairly close

Answer 1

Hint: $A_1P$ is the internal angle bisector of $\angle APA'$. $A_2P$ is the external angle bisector of $\angle APA'$.

Hence $\angle A_1 P A_2 = \frac{ 180^\circ} {2} = 90^\circ$.

Answer 2

The sum of the internal angle and the exterior angle is 180.

If you bisect internal angle and the exterior angle, then you've essentially divided 180 in half.