First, it is proved that AECF is a parallelogram. Conditions: AE=CF, AE//cf. There is AF=CE, and then it is proved that AG=CH. Conditions: angle BAE= angle DCF, angle GAE= angle HCF. (The former item can be obtained) So there is: angle package = angle DCH. Similarly, Angel ·ABG = Angel ·CDH. Then AB=CD. Available; AG=CH。 So GF=HE. And there is GF// he. So GFHE is a parallelogram. So EF and GH are equally divided.