Advanced Geometric Problem & Solution (Twice a week update) Now 4636 problems posted.
<Tri-2025-2238>, hint
Obviously I is te incenter of ABC, and since IP=IM=IN, triangle IPM is isosceles.
<BIC=90+A/2 and <MIC=90. Hence <MIB=<PIB=A/2, <IPM=90-A/2, Thus <API=90-C/2
Hence <IPA=<INC=90-C/2
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