Question

1如图,在角ABC中,BE、CF分别是∠B、∠C的平分线,求证：∠BPC等于90度家二分之一∠A.

Answer 1

证明：PB平分∠ABC；PC平分∠ACB,则：∠PBC=(1/2)∠ABC;∠PCB=(1/2)∠ACB.

故:∠BPC=180°-(∠PBC+∠PCB)=180°-(1/2)*(∠ABC+∠ACB)=180°-(1/2)*(180°-∠A)

=90°+(1/2)∠A.