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.

问题描述:

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.

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.