第一象限内有等腰三角形,已知顶点坐标,底边中点的坐标和底边长,求另两点坐标.设顶点A,顶点D,另两点B和C.已知AxAyDxDyBC求:Bx=By=Cx=Cy=设顶点A,底边中点D,另两点B和C.已知AxAyDxDyBC求:

Question

第一象限内有等腰三角形,已知顶点坐标,底边中点的坐标和底边长,求另两点坐标.
设顶点A,顶点D,另两点B和C.
已知 Ax Ay Dx Dy BC
求:
Bx =
By =
Cx =
Cy =
设顶点A,底边中点D,另两点B和C.
已知 Ax Ay Dx Dy BC
求:
Bx =
By =
Cx =
Cy =

Answer 1

复数法：
设复数DA为z=a+bi=r(cosθ+isinθ)
其中a=Ax-Dx b=Ay-Dy
则复数DB为DA以D为中心逆向旋转π/2
DB
=z(cosπ/2+isinπ/2)
=r[cos(θ+π/2)+isin(θ+π/2)]
=r[-sinθ+icosθ]
=-b+ai
复数DC为DA以D为中心正向旋转π/2
DC
=z(cos-π/2+isin-π/2)
=r[cos(θ-π/2)+isin(θ-π/2)]
=r[sinθ+icosθ]
=b-ai
∴DB=Dy-Ay+(Ax-Dx)i
DC=Ay-Dy+(Dx-Ax)i
∴复数B
=复数DB+复数D
=Dy-Ay+(Ax-Dx)i+Dx+Dyi
=Dx+Dy-Ay+(Ax-Dx+Dy)i
复数C
=复数DC+复数D
=Ay-Dy+(Dx-Ax)i+Dx+Dyi
=Dx-Dy+Ay+(Dx-Ax+Dy)i
坐标：
B(Dx+Dy-Ay,Ax-Dx+Dy)
C(Dx-Dy+Ay,Dx-Ax+Dy)
∴
Bx=Dx+Dy-Ay
By=Ax-Dx+Dy
Cx=Dx-Dy+Ay
Cy=Dx-Ax+Dy