> 数学 >
如图,在△ABC中,AD平分∠BAC交BC于D,BE⊥AC于E,交AD于F,求证:∠AFE=
1
2
(∠ABC+∠C).
人气:354 ℃ 时间:2019-10-17 02:40:31
解答
∵三角形内角和是180°,
∴∠BAC=180°-(∠ABC+∠C),
∵AD平分∠BAC交BC于D,
∴∠DCA=
1
2
∠BAC=90°-
1
2
(∠ABC+∠C),
∵BE⊥AC于E,
∴∠AFE=90°-∠FAE=90°-90°+
1
2
(∠ABC+∠C)=
1
2
(∠ABC+∠C).
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版