Let E be midpoint of the side AD of the square ABCD, O the circumcenter of triangle DEB. The circle (DEB) intersects the circle (E, EA) at D and G. Prove that 3BG=2OA.
My proof:
If P is the center of the square, then obviously 3AP=2OA, so we need to prove BG=PA ( * ). As O lies on AC (perpendicular bisector of BD, the circle O will also pass through F, midpoit of AB. PEAF is a square, thus EF=AP ( 1 ). As EG=DE=BF, BEGF is an isosceles trapezoid, wherefrom BG=EF; with (1), our claim (*) has been proved.
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