# Pythagoras and power of a point

Here is an attractive geometry problem that doesn’t require much angle chasing. A friend posed this problem to me, although I believe the original source was from somewhere on Brilliant.org (the problem may have been slightly adapted). Here is the question and my solution:
The point P is located within a circle Γ with centre O. Q lies on Γ such that ∠OPQ = 90. The tangent to Γ at Q meets the line OP at R. XY is a chord of circle Γ that passes through P. Prove that R, X, O and Y are concyclic.

Proof.
Let Γ cut PR internally at A, and let PO produced cut Γ again at B. Now let OQ = r and ∠POQ = x.

By angles in a triangle, ∠OQP = 90 – x.
Tangent and radius are perpendicular, so ∠PQR = x
By angles in a triangle, ∠PRQ = 90 – x
So triangles OPQ and PQR are similar by AAA similarity; hence, PQ/PO = PR/PQ
PQ² = PO × PR
Next, by intersecting chords theorem/power of a point theorem/whatever you call it, PX × PY = PA × PB
But PA = r – PO and PB = r + PO
So PA × PB = (r + PO)(r – PO) = r² – PO²
r² – PO² = PQ² by Pythagoras
Hence PO × PR = PQ² = PA × PB = PX × PY
PO × PR = PX × PY
So by the converse of the intersecting chords theorem/power of a point theorem/whatever you call it, R, X, O and Y are concyclic.