BMO1 2009 3


So I have drawn the intersection of circle (ABC) and CP at Q, which is located beyond segment CP. This can also be proven in the reverse order, when Q lies within CP.
First of all, note that <BAC = <BPC (properties of a parallelogram).
We also know that <BAC = <BQC (because ABCQ is cyclic).
Then triangle BPQ is isosceles and BP = BQ.

Now take a look at quadrilateral ABCQ. Since AB and CQ are parallel, the perpendicular bisector of AB passes through the circumcenter and is also the perpendicular bisector of CQ. Thus AC = BQ. (another way to prove this is to equate the angles of the two triangles: < QBA = <CAB = <ACQ = <CQB which follows from alternate interior angles).

If PQ = AC then PQ = AC = BQ = BP (proven above) which shows that triangle BPQ is equilateral. Then <BAC must be 60 degrees. These steps can be reversible so QED.

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