如图,AD是角BAC的平分线,活三角形ABC的边BC的中点M作AD的平行线,交AB于点E,交CA的延长线于点F,求...
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∵MF∥AD,AD位角平分线,故∠CAD=∠F=∠BAD=∠AEF,∴AE=AF

故AB/BE=BD/BM=AD/ME,CA/CF=CD/CM=AD/MF

把两个式子相加得AB/BE+CA/CF=BD/BM+CD/CM=AD=2

而AB=AE+BE,CA=CF-AF,故AB/BE+CA/CF=1+AE/BE+1-AF/CF=2

得AE/BE=AF/CF,故BE=CF