已知：△ABC，如图，若P点是∠ABC和∠ACB的角平分线的交点，求证：∠P=90°+12∠A．

Question

已知：△ABC，如图，若P点是∠ABC和∠ACB的角平分线的交点，求证：∠P=90°+(1)/(2)∠A．

Answer 1

证明：∵P点是∠ABC和∠ACB的角平分线的交点，
∴∠PBC+∠PCB=(1)/(2)（∠ABC+∠ACB）=(1)/(2)（180°-∠A）=90°-(1)/(2)∠A．
∴∠P=180°-(1)/(2)（∠ABC+∠ACB）=180°-90°+(1)/(2)∠A=90°+(1)/(2)∠A．