IBM168
幼苗
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设A1C1与BC相交于O点,过C点做CD垂直于AB交AB于D点,过O点做OE垂直于A1B交A1B于E点,则AB×CD/2=2×A1B×OE/2,即OE/CD=AB/2A1B,
∵△ABC∽△A1BO,∴A1B/AB=BO/BC,
又∵△BOE∽△BCD,∴OE/CD=BO/BC
∴A1B/AB=OE/CD,即A1B/AB=AB/2A1B,求得AB=√2A1B,AA1=AB-A1B=(√2-1)A1B.
注:你那个AB=?,没写明白,自己往里填数值.
1年前
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