在正方形abcd中 点E和点F是边AD和DC的中点 连接BF CE 求证AB=AP图片在http://hi.baidu.com/kl2797/album/item/3e07a4c800717d2a00e928a5.html 答案:延长CE交AB延长线于G∵CF=ED.∠BCF=∠EDC=90°.BC=CD∴△BCF≌△CDE∴∠CBF=∠ECD∵∠CBF+∠BFC=90°∴∠ECD+∠BFC=90°∴BF⊥EC在△GEA和△EDF中,易得两三角形全等所以PA是RT△PGB斜边中线∴PA=AB 第二题如下图所示,在锐角△ABC中,AB=4√2,∠BAC=45°,∠BAC的平分线交BC于点D,M、N分别是AD和AB上的动点,则BM+MN的最小值是?
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如图,在AC线上取一点E使AE=AB.自E点向AB作垂线交AD于F、交AB于G;自M点向EG作垂线垂足为H.连接BE、BF、EM.ABE为等腰三角形,AD为∠A的平分线,则AD必为BE的垂直平分线,故BF=EF、 BM=EM。当动点M不与F点重合时,BM=EM>EH、 MN≥HG,则BM+MN>EH+HG=EG;当M点与F点重合、N点与G点重合时,BM+MN=EF+FG=EG.故知:BM+MN≥EG.AEG为直角等腰三角形,所以EG=AE/√2=4√2/√2=4.得BM+MN≥4,即BM+MN的最小值是4.
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