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).
如图,在△ABC中,AD平分∠BAC交BC于D,BE⊥AC于E,交AD于F,求证:∠AFE=
(∠ABC+∠C).
数学人气:301 ℃时间:2019-08-20 19:29:15
优质解答
∵三角形内角和是180°,
∴∠BAC=180°-(∠ABC+∠C),
∵AD平分∠BAC交BC于D,
∴∠DCA=
∠BAC=90°-
(∠ABC+∠C),
∵BE⊥AC于E,
∴∠AFE=90°-∠FAE=90°-90°+
(∠ABC+∠C)=
(∠ABC+∠C).
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