Let $ABCD$ be a quadrilateral inscribed in a circle such that $\angle ABC=60^{\circ}.$ a) Prove that if $BC=CD$ then $AB= CD+DA.$ b) Is it true that if $AB= CD+DA$ then $BC=CD$?
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Tags: geometry, Angle Chasing, Argentina
Let $ABCD$ be a quadrilateral inscribed in a circle such that $\angle ABC=60^{\circ}.$ a) Prove that if $BC=CD$ then $AB= CD+DA.$ b) Is it true that if $AB= CD+DA$ then $BC=CD$?