MathDB
Show that m(BCD)=90

Source: Turkey Junior National Olympiad 2012 P2

December 12, 2012
geometrycircumcircleratiotrapezoidsymmetrygeometric transformationreflection

Problem Statement

In a convex quadrilateral ABCDABCD, the diagonals are perpendicular to each other and they intersect at EE. Let PP be a point on the side ADAD which is different from AA such that PE=EC.PE=EC. The circumcircle of triangle BCDBCD intersects the side ADAD at QQ where QQ is also different from AA. The circle, passing through AA and tangent to line EPEP at PP, intersects the line segment ACAC at RR. If the points B,R,QB, R, Q are concurrent then show that ∠BCD=90∘\angle BCD=90^{\circ}.
Back to ProblemsView on AoPS