Let $A$ be any point outside the circle $\omega$, let $A^*$ be its inverse around $\omega$, and $P$ is any variable point on $\omega$. Prove that the ratio $PA$/$PA^*$ is constant, and hence find its value in terms of $P$ , $A^*$, and $A$.
Upto which I could do, but can't understand the last line of my solution:
Because A' is inverse of A around ω we can write: OA * OA' = r² ⇔ OA / r = r / OA' ⇔ OA / OP = OP / OA' (1)
Because ∡α has the sides proportional in △OAP and △OPA' then triangles are similar. (Two triangles are similar if the corresponding lengths of two sides are proportional and the included angles are congruent (SAS).) We can add to (1) the ratio: PA / PA' = OA / OP = OP / OA' (2). Let k be the common value of those ratios. We'll prove that k is constant. Indeed, from properties of proportions we have: If a/b = c /d then we can add the ratio (a - c) / (b - d) to that proportion and so we can write: a/b = c/d = (a - c) / (b - d). So applying this property to the last proportion in (2) we have: k = OA / OP = OP / OA' = (OA - OP) / (OP - OA'). But OP = OB (where B = ω ∩ OA) and we have: OA - OP = OA - OB = AB and OP - OA' = OB - OA' = A'B So we have: k = PA / PA' = AB / A'B = const, qed.
Observation: The last proportion says that PB is the angle bisector in △APA'. The angle bisector theorem is concerned with the relative lengths of the two segments that a triangle's side is divided into by a line that bisects the opposite angle. It equates their relative lengths to the relative lengths of the other two sides of the triangle.