MathLabs

Problem 3

Let ABCABC be an acute triangle with ∠BAC=30∘\angle BAC=30^\circ. The internal and external bisectors of ∠ABC\angle ABC meet line ACAC at B1,B2B_1,B_2, and those of ∠ACB\angle ACB meet line ABAB at C1,C2C_1,C_2. The circles with diameters B1B2B_1B_2 and C1C2C_1C_2 meet inside ABCABC at PP. Prove that ∠BPC=90∘\angle BPC=90^\circ.
Step 4 of 6: Cyclic angle
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∠CPE=∠CDE\angle CPE=\angle CDE
Detailed analysis

Since C,D,P,EC,D,P,E are cyclic, ∠CPE=∠CDE\angle CPE=\angle CDE; similarly ∠FPB=∠FDB\angle FPB=\angle FDB.