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 5 of 6: Angle sum
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∠BPC=90∘\angle BPC=90^\circ
Detailed analysis

Using DEFDEF equilateral and the angle sum in the pedal configuration, the official angle chase gives ∠BPC=360∘−(∠CDE+∠FDB+∠EPF)=90∘\angle BPC=360^\circ-(\angle CDE+\angle FDB+\angle EPF)=90^\circ.