来自盛军强的问题
(2011•黄浦区二模)如图,AB与CD相交于点O,AD∥BC,AD:BC=1:3,AB=10,则AO的长是5252.
(2011•黄浦区二模)如图,AB与CD相交于点O,AD∥BC,AD:BC=1:3,AB=10,则AO的长是52
52
.
1回答
2020-05-20 17:54
(2011•黄浦区二模)如图,AB与CD相交于点O,AD∥BC,AD:BC=1:3,AB=10,则AO的长是5252.
(2011•黄浦区二模)如图,AB与CD相交于点O,AD∥BC,AD:BC=1:3,AB=10,则AO的长是52
52
.
∵AD∥BC,
∴△AOD∽△BOC,
∴AD:BC=OA:OB=1:3,
∵AB=10.OA+OB=AB,
∴AO=52