如图,点G、E、F分别在平行四边形ABCD的边AD、DC和BC上,DG=DC,CE=CF...

发布网友发布时间：2024-07-05 03:25

共4个回答

热心网友时间：2024-07-16 12:27

解法一：

1、从G点作DC的平行线GH，得到平形四边形GHCD。2、因为DC=DG,平行四边形GHCD就是个等边的平行四边形。所以呢GP就是平分角DCB的，所以解难ECP=角FCP，又因为EC=FC,三角形ECP和三角形FCP又有一条公共边，所以呢EP=FP了.

解法二：

1、因为DA平行于CB，所以角DGC=角GCB.

又因为DG=DC,所以角DGC=角DCG.

所以角DCG=角GCB,所以角ECP=角FCP.

又因为边EC=边FC，CP是三角形ECP和三角形FCP的公共边，（两边夹一角相等）所以边EP=边FP

热心网友时间：2024-07-16 12:26

证明：∵四边形ABCD是平行四边形，
∴AD∥BC，
∴∠DGC=∠GCB（两直线平行，内错角相等），
∵DG=DC，
∴∠DGC=∠DCG，
∴∠DCG=∠GCB，
∵∠DCG+∠DCP=180°，∠GCB+∠FCP=180°，
∴∠DCP=∠FCP，
∵在△PCF和△PCE中
CE=CF∠FCP=∠ECPCP=CP，
∴△PCF≌△PCE（SAS），
∴PF=PE．

热心网友时间：2024-07-16 12:28

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热心网友时间：2024-07-16 12:23

给个图，要不然不知道