Two circles, A and B, intersect at X and Y. P is a variable point on A. PX meets B again at Q, and QY meets A again at R.

**The length of the chord RP is constant, irrespective of the position of P on A.
**

Proof.

To prove this, I will fix P and consider the new point P’, defining Q’ and R’ in the same way as before. I will show RP = R’P’. Then, wherever you move P’, the length of R’P’ will not change as P is fixed.

There are two cases to consider: whether P’ is on the same side of XY as Q’ or not.

(There are actually other cases, when the circles overlap to the extent that R and P are on opposite sides of XY, but the methods of proof are very similar. See the end of this post for a diagram of one such configuration)

**Case 1: **P’ is on the opposite side of XY as Q’

Let ∠PXP’ = θ.

∠QXQ’ = θ (vertically opposite)

∠QYQ’ = θ (same segment)

∠RYR’ = θ (vertically opposite)

Since arcs PP’ and RR’ subtend equal angles, arc length PP’ = arc length RR’

So arc length RP = arc length RR’ + arc length R’P = arc length PP’ + arc length R’P = arc length R’P’

So chord length RP = chord length R’P’

**Case 2: **P’ is on the same side of XY as Q’

Let ∠PRP’ = θ

∠PXQ’ = 180 – θ (opposite angle in cyclic quadrilateral PRP’X)

∠QXQ’ = θ (straight line)

∠QYQ’ = θ (same segment)

∠RYR’ = θ (vertically opposite)

So arcs PP’ and RR’ subtend equal angles.

The proof is finished off as before.

By the way, here is one of the other cases:

There are other cases but the methods of proof are virtually identical to either of the two given here.