P' is the inverse of P with respect to the circle c and M is a point of a circle c. Line through M and P intersects with c at A and line through M and P' intersects with c at B. Prove that AB is orthogonal to PP'.
I think I'm supposed to invert lines through MP , MP' and AB. Line through PP' will stay the same after inversion, line through AB will become a circle around AB0, line through MP will become a circle around MAOP' and line through MP' will become a circle around MBOP.
I'm obviously supposed to prove that line through PP' contains a center of circle around ABO and I suppose I should somehow use cyclic quadrilaterals MBOP and MAOP' but I've been doing this problem on and of for couple days and nothing comes to mind
Is this a right aprroach? Any tips?
Best Answer
Disclaimer: this is not, strictly speaking, a solution to this question.
Indeed, I haven't found a solution using inversion but this answer calls the attention of the OP (and further readers) that this issue comes from a remarkable configuration well described here:
https://inversivegeometry.quora.com/Points-Part-1-2
(I have borrowed one of its figures: see below ; I have just changed the names of the vertices).
Let us first describe the logic of construction of the following figure which is somewhat inverse (pun intended) of the construction of the initial issue: we start from a circle (c) with radius $OE=r$ with 2 symmetrical points $A,B \in (c)$ with $AB \perp EO$. We choose a point $M \in (c)$, we draw lines $MA$ and $MB$ intersecting line $OE$ in $P$ and $P'$ resp. We want to show that $P$ and $P'$ are mutual inverses.
The central property of this figure is that it contains three similar triangles
$$BOP' \sim P'MP \sim AOP \tag{1}$$
(these similarities being obtained by angle chasing ; the starting property being $\angle EOB = \angle AMB = \alpha$ due to the inscribed angle property.
As a consequence of (1):
$$\frac{OP}{OA}=\frac{OB}{OP'} \ \iff \ OP.OP'=r^2$$
proving that $P$ and $P'$ are indeed mutual inverses.