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 3 of 6: Equilateral pedal triangle
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EF=FD=DEEF=FD=DE
Detailed analysis

Cyclically, FD=PBsin⁡BFD=PB\sin B and DE=PCsin⁡CDE=PC\sin C. The equalities from step 1 therefore imply DE=EF=FDDE=EF=FD.