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right angle wanted, <APC=180^o-<ABC,BC = AP, AK=KB+PC (2018 Kyiv City MO 7.4)

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September 11, 2020
geometryright angleequal segmentsangles

Problem Statement

Inside the triangle ABCABC , the point PP is selected so that BC=APBC = AP and ∠APC=180∘−∠ABC\angle APC = 180 {} ^ \circ - \angle ABC . On the side ABAB there is a point KK , for which AK=KB+PCAK = KB + PC . Prove that ∠AKC=90∘\angle AKC = 90 {} ^ \circ .
(Danilo Hilko)
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