证明:
∵∠DBC=180-∠ABC,BP平分∠DBC
∴∠PBC=∠DBC/2=(180-∠ABC)/2=90-∠ABC/2
∵∠ECB=180-∠ACB,CP平分∠ECB
∴∠PCB=∠ECB/2=(180-∠ACB)/2=90-∠ACB/2
∴∠BPC=180-(∠PBC+∠PCB)
=180-(90-∠ABC/2+90-∠ACB/2)
=(∠ABC+∠ACB)/2
=(180-∠A)/2
=90-∠A/2
证明:
∵∠DBC=180-∠ABC,BP平分∠DBC
∴∠PBC=∠DBC/2=(180-∠ABC)/2=90-∠ABC/2
∵∠ECB=180-∠ACB,CP平分∠ECB
∴∠PCB=∠ECB/2=(180-∠ACB)/2=90-∠ACB/2
∴∠BPC=180-(∠PBC+∠PCB)
=180-(90-∠ABC/2+90-∠ACB/2)
=(∠ABC+∠ACB)/2
=(180-∠A)/2
=90-∠A/2