MathDB

3

Part of 2002 Iran MO (2nd round)

Problems(1)

nice problem(due hard)

Source: iranian second round (2002)

5/10/2005
In a convex quadrilateral ABCDABCD with ∠ABC=∠ADC=135∘\angle ABC = \angle ADC = 135^\circ, points MM and NN are taken on the rays ABAB and ADAD respectively such that ∠MCD=∠NCB=90∘\angle MCD = \angle NCB = 90^\circ. The circumcircles of triangles AMNAMN and ABDABD intersect at AA and KK. Prove that AK⊥KC.AK \perp KC.
geometrycircumcircleparallelogramgeometry solved