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equal angles when <C = <A + 90^o

Source: IGO 2014 Junior 4

July 22, 2018
geometryequal angles

Problem Statement

In a triangle ABC we have ∠C=∠A+90o\angle C = \angle A + 90^o. The point DD on the continuation of BCBC is given such that AC=ADAC = AD. A point EE in the side of BCBC in which AA doesn’t lie is chosen such that ∠EBC=∠A,∠EDC=12∠A\angle EBC = \angle A, \angle EDC = \frac{1}{2} \angle A . Prove that ∠CED=∠ABC\angle CED = \angle ABC.
by Morteza Saghafian
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