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 1 of 6: Apolonius circles
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PAsin⁡A=PBsin⁡B=PCsin⁡CPA\sin A=PB\sin B=PC\sin C
Detailed analysis

Because PP lies on the circle with diameter B1B2B_1B_2, the Apollonius ratio gives PA/PC=BA/BC=sin⁡C/sin⁡APA/PC=BA/BC=\sin C/\sin A, hence PAsin⁡A=PCsin⁡CPA\sin A=PC\sin C. The other circle similarly gives PAsin⁡A=PBsin⁡BPA\sin A=PB\sin B.