三角形ABC中,D是BC的中点,AE平分∠BAC,BE⊥AE宇E点,若AB=5,AC=7,求ED
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延长BE,交AC于点或正陵F.
在△AEF和△AEB中,∠AEF = 90°= ∠AEB ,AE为公共边,∠EAF = ∠EAB ,
所以,△AEF ≌ △AEB ,可得：EF = EB ,AF = AB = 5 .
因为清孙,EF = EB ,CD = BD ,可得：DE是△BCF的中衫戚位线,
所以,DE = (1/2)CF = (1/2)(AC-AF) = 1 .
在△AEF和△AEB中,∠AEF = 90°= ∠AEB ,AE为公共边,∠EAF = ∠EAB ,
所以,△AEF ≌ △AEB ,可得：EF = EB ,AF = AB = 5 .
因为清孙,EF = EB ,CD = BD ,可得：DE是△BCF的中衫戚位线,
所以,DE = (1/2)CF = (1/2)(AC-AF) = 1 .
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