已知梯形ABCD中 AB//CD 对角线AD,BC相交于O MN过O平行AB交AC于M BD于N MN=1 求1/AB
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已知梯形ABCD中 AB//CD 对角线AD,BC相交于O MN过O平行AB交AC于M BD于N MN=1 求1/AB + 1/CD的值
MN平行AB,则OM/AB=CM/CA; ON/AB=DN/DB=CM/CA.
故:OM/AB+ON/AB=CM/CA+CM/CA,即:(OM+ON)/AB=2CM/CA,1/AB=2CM/CA.----------(1)
同理可证:MO/CD=AM/CA; ON/CD=BN/BD=AM/CA.
则MO/CD+ON/CD=AM/CA+AM/CA,即(OM+ON)/CD=2AM/CA,1/CD=2AM/CA.-------------(2)
(1)+(2),得:1/AB+1/CD=(2CM+2AM)/CA=2CA/CA=2.
故:OM/AB+ON/AB=CM/CA+CM/CA,即:(OM+ON)/AB=2CM/CA,1/AB=2CM/CA.----------(1)
同理可证:MO/CD=AM/CA; ON/CD=BN/BD=AM/CA.
则MO/CD+ON/CD=AM/CA+AM/CA,即(OM+ON)/CD=2AM/CA,1/CD=2AM/CA.-------------(2)
(1)+(2),得:1/AB+1/CD=(2CM+2AM)/CA=2CA/CA=2.
已知梯形ABCD中 AB//CD 对角线AD,BC相交于O MN过O平行AB交AC于M BD于N MN=1 求1/AB+
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