在三角形abc中,AB=AC,AE垂直BC于E,在BC上取一点D,使CD=CA,连接AD,若AD=DB,则∠DAE=多少
问题描述：
在三角形abc中,AB=AC,AE垂直BC于E,在BC上取一点D,使CD=CA,连接AD,若AD=DB,则∠DAE=多少
答
因为 AB=AC
所以 ∠ABC = ∠ACB， 为等腰三角形
三角形内角之和 ∠ABC + ∠ACB + ∠BAC = 180
=> ∠BAC = 180 - 2∠ABC
因为AE垂直BC于E， 且ABC为等腰三角形
所以 ∠BAE = ∠BAC/2 = 90 - ∠ABC
∠DAE = ∠BAE - ∠BAD
因为AD=DB，所以 ∠BAD = ∠ABC
=> ∠DAE = 90 - ∠ABC - ∠ABC = 90 - 2∠ABC
在三角形ADC内
因为CD=CA，所以 ∠DAC = ∠ADC = (180 - ∠ACB)/2
=> ∠DAC = 90 - ∠ABC/2
在直角三角形ADE内
∠ADE = 90 - ∠DAE
而 ∠ADE = ∠ADC = 90 - ∠ABC/2
=> ∠DAE = ∠ABC/2
而同时 ∠DAE = 90 - 2∠ABC
=> (5/2)∠ABC = 90
=> ∠ABC = 36
则 ∠DAE = ∠ABC/2 = 36/2 = 18度
答
18°