Omni-MATH / In a right angled-triangle ABC, ∠ACB=90^(o). Its incircle O meets BC, AC, AB at D,E,F respectively. AD cuts O at P. If ∠BPC=90^(o), prove AE+AP=PD.

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In a right angled-triangle ABC, ∠ACB=90^(o). Its incircle O meets BC, AC, AB at D,E,F respectively. AD cuts O at P. If ∠BPC=90^(o), prove AE+AP=PD.

In a right angled-triangle $ABC$, $\angle{ACB} = 90^o$. Its incircle $O$ meets $BC$, $AC$, $AB$ at $D$,$E$,$F$ respectively. $AD$ cuts $O$ at $P$. If $\angle{BPC} = 90^o$, prove $AE + AP = PD$.

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task_id

e8387dd5-3fea-51e7-ad32-dc86cd5db0bf

task_key

test--e8387dd5-3fea-51e7-ad32-dc86cd5db0bf

task_revision_id

4

config / split

/ test

Official source

initial import