证明:
∵∠FBC =∠A+∠ACB,∠EBC=∠A +∠ABC
∴∠FBC+∠ECP=∠A+∠ACB+∠A +∠ABC
∵BP和CP是外角平分线
∴∠PBC+∠PCB=1/2(∠A+∠ACB+∠A +∠ABC)=∠A +1/2(180°-∠A)=90°+1/2∠A
∴∠BPC=180°-(90°+1/2∠A)=90°-1/2∠A
证明:
∵∠FBC =∠A+∠ACB,∠EBC=∠A +∠ABC
∴∠FBC+∠ECP=∠A+∠ACB+∠A +∠ABC
∵BP和CP是外角平分线
∴∠PBC+∠PCB=1/2(∠A+∠ACB+∠A +∠ABC)=∠A +1/2(180°-∠A)=90°+1/2∠A
∴∠BPC=180°-(90°+1/2∠A)=90°-1/2∠A