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A funny angle equality in a rigid square configuration

Source: Czech-Polish-Slovak Junior Match 2024, I-3

May 29, 2024
geometrygeometry proposed

Problem Statement

Let ABCDABCD be a convex quadrilateral with AB=BD=DCAB=BD=DC and AB⊥BD⊥DCAB \perp BD \perp DC. Let MM be the midpoint of segment BCBC. Show that ∠BAM+∠DCA=45∘\angle BAM+\angle DCA=45^\circ.
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