在△ABC中,∠A=40°(1)如图1,若两内角∠ABC、∠ACB的角平分线交于点P,则∠P=______,∠A与∠P之间的

2025-05-14 03:25:24
推荐回答(1个)
回答1:

(1)∠ABC+∠C=180°-∠A=180°-40°=140°

1
2
(∠ABC+∠C)=
1
2
×140°=70°,
∴∠P=180°-
1
2
(∠ABC+∠C)=110°.
∠A与∠P之间的数量关系是∠P=90°+
1
2
∠A;

(2)∵
1
2
∠ACE=
1
2
∠ABC+∠P,
1
2
(∠A+∠ABC)=
1
2
∠ABC+∠P,
1
2
(40°+∠ABC)=
1
2
∠ABC+∠P,
∴∠P=20°.
∠A与∠P之间的数量关系是∠P=
1
2
∠A;

(3)∵∠EBC=∠A+∠ACB,∠BCF=∠A+∠ABC,
∴∠EBC+∠BCF=∠A+∠ACB+∠ABC+∠A=180°+∠A,
∴∠PBC+∠PCB=90°+
1
2
∠A.
又∵∠PBC+∠PCB+∠P=180°,
∴90°+
1
2
∠A+∠P=180°,即∠P=90°-
1
2
∠A.