Two circles Γ₁ and Γ₂ intersect at points P and Q.

Prove that there are infinitely many cyclic quadrilaterals
ABCD ( A and B on Γ₁ and C and D on Γ₂ ) such that

         AC∩BD is in the set {P,Q}.

For any point, A, chosen on Circle1, points B, C and D can be constructed
so that ABCD is cyclic and as specified.
However, to keep the labelling unique, let the tangent to Circle2 at P
cross Circle1 at X; then choose A to lie on the arc PX, of Circle1, that does
not include Q (still an infinite number of possible positions for A!).

            Let AP cut Circle2 at C.
            Let the reflection of AQ in PQ cut Circle2 at D.
            Let DP cut Circle1 at B.
            ABCD is then cyclic.
Proof    
/PQA = /PQD     by symmetry about PQ
/PBA and /PQA are equal if in same segment, otherwise complementary
/PCD and /PQD are equal if in same segment, otherwise complementary
These conditions correspond, so in both cases /PBA = /PCD (/DBA = /ACD).

Since /DBA and /ACD are both subtended by AD, a circle passing through
A, D and C must also pass through B. So ABCD is cyclic.

Posted by Harry on 2013-10-26 11:42:12