如图:AB=CD,AE=DF,CE=FB.求证:AE∥DF
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如图:AB=CD,AE=DF,CE=FB.求证:AE∥DF
运用三角形全等的原理.
∵CE=FB
∴CE+EF=FB+EF
即CF=BE
又∵AB=CD AE=DF
∴△AEB≡(全等于)△DFC(SSS)
∴∠AEB=∠DFC
即AE∥DF
答:
本题得证
因为CE=BF 得CE+EF=BF+EF
又因为AB=CD AE=DF CF=BE
得三角形ABE全等三角形CFD
所以角AEB=角CFD
得AE平行于DF
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