题目内容
如图,在△ABC中,∠BAD=∠DAC,BE⊥AC于E,交AD于F.试说明∠AFE=
(∠ABC+∠C).
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证明:∵∠BAD=∠DAC,
∴∠DAC=
∠BAC,
∵∠BAC+∠ABC+∠C=180°,
∴∠BAC=180°-(∠ABC+∠C),
∴∠DAC=
[180°-(∠ABC+∠C)],
=90°-
(∠ABC+∠C),
∵BE⊥AC,
∴∠AEB=90°,
∴∠AFE+∠DAC=90°,
∴∠AFE=90°-∠DAC=90°-90°+
(∠ABC+∠C),
=
(∠ABC+∠C).
∴∠DAC=
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∵∠BAC+∠ABC+∠C=180°,
∴∠BAC=180°-(∠ABC+∠C),
∴∠DAC=
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=90°-
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∵BE⊥AC,
∴∠AEB=90°,
∴∠AFE+∠DAC=90°,
∴∠AFE=90°-∠DAC=90°-90°+
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=
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