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Part of 2007 Poland - Second Round

Problems(2)

Polish MO 2007 Second Round Day 1, problem 2

Source:

2/28/2007
ABCDEABCDE is a convex pentagon and: BC=CD,      DE=EA,      ∠BCD=∠DEA=90∘BC=CD, \;\;\; DE=EA, \;\;\; \angle BCD=\angle DEA=90^{\circ} Prove, that it is possible to build a triangle from segments ACAC, CECE, EBEB. Find the value of its angles if ∠ACE=α\angle ACE=\alpha and ∠BEC=β\angle BEC=\beta.
two rhombi

Source: Polish 2007 2nd Round

3/2/2007
We are given a cyclic quadrilateral ABCD   AB\not=CD. Quadrilaterals AKDLAKDL and CMBNCMBN are rhombuses with equal sides. Prove, that KLMNKLMN is cyclic
geometrycyclic quadrilateralpower of a pointradical axisgeometry unsolved