# Circle of Apollonius proof 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

• Hint: Half of the angle on a straight line is a right angle. Commented Sep 26, 2013 at 4:59
• @CalvinLin I still don't see it. I dont think we get half an angle on a striaght line, we get half an arbitrary interior angle, and half an exterior angle that supplements it, meaning both angles (and their halfs) are arbitrary, no? Commented Sep 26, 2013 at 5:07

Hint: $A_1P$ is the internal angle bisector of $\angle APA'$. $A_2P$ is the external angle bisector of $\angle APA'$.
Hence $\angle A_1 P A_2 = \frac{ 180^\circ} {2} = 90^\circ$.
• Why does it equal $\frac{180}{2}$? Commented Sep 26, 2013 at 5:14
• @Dgrin91 Work it out. Let $\angle APA' = \alpha$. Write out all the angles in terms of $\alpha$. Commented Sep 26, 2013 at 5:15