Question

一道难题，希望帮助一下？

用初中的方法

Answer 1

设AB=x,BD=y,∠B=90°，CD=8,
tanC=AB/BC=x/(y+8),
∠BAD=2∠C,
所以tan2C=y/x=[2x/(y+8)]/[1-x^2/(y+8)^2]=2x(y+8)/[(y+8)^2-x^2],
去分母得y[(y+8)^2-x^2]=2x^2*(y+8),
整理得y(y+8)^2=x^2*(3y+16),
x^2=y(y+8)^2/(3y+16),①
由勾股定理，AC^2=x^2+(y+8)^2=144.②
把①代入②*（3y+16）,得y(y+8)^2+(3y+16)(y+8)^2=144(3y+16),
整理得(y+4)(y+8)^2=36(3y+16),
(y+4)(y^2+16y+64)=108y+576,
y^3+16y^2+64y
........+4y^2+64y+256
..................-108y-576=0,
y^3+20y^2+20y-320=0,
解得y≈3.30142383，或y^2+23.30142 y+ 96.92788 ≈0，无正根。
代入①，x^2≈16.27781931，
AB=x≈4.03457796.

Answer 2

望采纳

Answer 3

想你了半天没想出来。还问老师吧