如图所示,梯形ABCD中,AB∥CD,两对角线交于点O,过点O做两底的平行线交两腰AD,BC于M、N.求证1/AB+1/CD=2/MN
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![如图所示,梯形ABCD中,AB∥CD,两对角线交于点O,过点O做两底的平行线交两腰AD,BC于M、N.求证1/AB+1/CD=2/MN](/uploads/image/z/10268956-28-6.jpg?t=%E5%A6%82%E5%9B%BE%E6%89%80%E7%A4%BA%2C%E6%A2%AF%E5%BD%A2ABCD%E4%B8%AD%2CAB%E2%88%A5CD%2C%E4%B8%A4%E5%AF%B9%E8%A7%92%E7%BA%BF%E4%BA%A4%E4%BA%8E%E7%82%B9O%2C%E8%BF%87%E7%82%B9O%E5%81%9A%E4%B8%A4%E5%BA%95%E7%9A%84%E5%B9%B3%E8%A1%8C%E7%BA%BF%E4%BA%A4%E4%B8%A4%E8%85%B0AD%2CBC%E4%BA%8EM%E3%80%81N.%E6%B1%82%E8%AF%811%2FAB%2B1%2FCD%3D2%2FMN)
如图所示,梯形ABCD中,AB∥CD,两对角线交于点O,过点O做两底的平行线交两腰AD,BC于M、N.求证1/AB+1/CD=2/MN
如图所示,梯形ABCD中,AB∥CD,两对角线交于点O,过点O做两底的平行线交两腰AD,BC于M、N.求证1/AB+1/CD=2/MN
如图所示,梯形ABCD中,AB∥CD,两对角线交于点O,过点O做两底的平行线交两腰AD,BC于M、N.求证1/AB+1/CD=2/MN
证明:因为MN平行BC平行AD
所以OM/AB=OB/BD
ON/BC=OD/BD
所以OM/AB+ON/BC=(OB+OD)/BD=BD/BD=1 (1)
同理可证:ON/AB=OC/AC
OM/CD=OA/AC
所以ON/AB+OM/CD=(OC+OA)/AC=AC/AC=1 (2)
(1)+(2)
(ON+OM)/AB+(ON+OM)/CD)=2
因为OM+ON=MN
所以MN/AB+MN/CD=2
所以1/AB+1/CD=2/MN
证明:∵AB∥MN∥CD.
∴MO/AB=DO/DB;ON/CD=BO/DB.
则MO/AB+ON/CD=(DO+BO)/DB=1;
同理可证:MO/CD+ON/AB=1.
∴(MO/AB+ON/AB)+(MO/CD+ON/CD)=2.
即MN/AB+MN/CD=2.
∴1/AB+1/CD=2/MN. ----------------(即上式两边同除以MN).
证明:∵AB∥MN∥CD.
∴MO/AB=DO/DB;ON/CD=BO/DB.
则MO/AB+ON/CD=(DO+BO)/DB=1;
同理可证:MO/CD+ON/AB=1.
∴(MO/AB+ON/AB)+(MO/CD+ON/CD)=2.
即MN/AB+MN/CD=2.
∴1/AB+1/CD=2/MN. ----------------(即上式两边同除以MN). ∵OM+ON=MN
∴MN/AB+MN/CD=2
∴1/AB+1/CD=2/MN