梯形ABCD中,AB∥DC,对角线AC,BD交与O,过O作MN∥DC交AD于M,交BC于N 求证：OM/AB+ON/BC的值AB是_其他解答

梯形ABCD中,AB∥DC,对角线AC,BD交与O,过O作MN∥DC交AD于M,交BC于N 求证：OM/AB+ON/BC的值
AB是上底，DC是下底，A,D都在左边

人气：477 ℃　时间：2020-09-14 01:35:24

解答

∵MN∥DC,AB∥DC
∴MN∥AB
∴∠DAB=∠DMO,∠DBA=∠DOM
∴ΔABD∽ΔMOD
OM/AB=MD/AD
同理ON/BC=BN/BC
由AB∥MN∥CD可得BN/BC=AM/AD
∴OM/AB+ON/BC=MD/AD+AM/AD=AD/AD=1OM/AB+ON/BC=MD/AD+AM/AD，请问这是整么变的啊，没看懂ON/BC=BN/BC(已证)BN/BC=AM/AD(已证)所以ON/BC=AM/AD (1)而OM/AB=MD/AD (已证)(2)（1）+（2）可得OM/AB+ON/BC=MD/AD+AM/AD