**Two circles, X and Y, intersect at two points A and B. One line is drawn through X, cutting it at two points, P and Q; another line, not parallel to the first, is drawn through Y, cutting it at two points, R and S. These two lines meet at a point M. Then, M lies on AB if and only if P, Q, R and S are concyclic.**

Proof:

As this is an ‘if and only if’ result, we have two jobs.

**1. ****If M lies on AB, then ****P, Q, R and S are concyclic.**

**2. If P, Q, R and S are concyclic, then M lies on AB.**

Proof of 1.

By applying intersecting secant theorem/secant-secant theorem/power of a point theorem to X, MP ∙ MQ = MA ∙ MB

(We use intersecting chords theorem instead of intersecting secant theorem if M lies between A and B – if we use power of a point there is no need to consider different cases as power of a point uses directed lengths.)

By applying the same to Y, MR ∙ MS = MA ∙ MB

Hence MR ∙ MS = MP ∙ MQ

Hence by the converse of intersecting secant theorem, PQSR is cyclic.

Proof of 2.

We are going to use statement 1 to prove statement 2. We will follow a proof by contradiction.

Suppose P, Q, R and S are concyclic, but M does not lie on AB. Then PQ must intersect AB at a second point N.

(As you can see, ‘circle’ PQSR is a very perverse circle indeed – kind of egg shaped)

Let NS intersect circle Y at L. Then, by statement 1, PQSL is cyclic.

Saying that PQSR and PQSL are cyclic is equivalent to saying that the circumcircle of PQS passes through L and R.

But if this were true, then this circle would cut circle Y in 3 places: L, R and S; and no two circles can intersect in 3 places.

Hence we obtain a contradiction.

Therefore, if PQSR is cyclic, M lies on AB.

The source of this result is Geoff Smith, who showed and proved to me the first of these two statements at a UKMT olympiad training camp. (Therefore the first one is definitely correct – I worked the second one out myself, so for that I can’t say the same XD)

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