好像图形是不固定，但角ADB的大小不会变

一个三角解法:

设BD=CD=AE=a, 又用A,D,C表示三角形ADC中的三个角
  即 A=∠DAC, D=∠ADC, C=∠ACD
则 BC = 2a
∠BFC = 2∠EAF =2A
∠CBF = (180-∠BFC-C)=(180-2A-C)
CF/BC = sin∠CBF/sin∠BFC = sin(180-2A-C)/sin(2A)
      = sin(2A+C)/sin(2A)
CF = 2a * sin(2A+C)/sin(2A)
因 AC=DC * sinD/sinA =a * sinD/sinA
因此 AF=AC-CF =a*(sinD/sinA - 2sin(2A+C)/sin(2A))
AE = 2AF*cosA = 2cosA * a * (sinD/sinA - 2sin(2A+C)/sin(2A))
              = a*(2cosA*sinD/sinA - 4cosA*sin(2A+C)/sin(2A))
因AE=BD=a, 知
   2cosA*sinD/sinA - 4cosA*sin(2A+C)/sin(2A) = 1
   2cosA*sinD*2cosA - 4cosA*sin(2A+C) = 2sinA*cosA
   2sinD*cosA - 2sin(2A+C) = sinA
   2sinD*cosA - 2sin(2A+180-A-D) = sinA
   2sinD*cosA - 2sin(D-A) = sinA
   2sinD*cosA - 2sinD*cosA + 2cosD*sinA = sinA
   2cosD*sinA = sinA
   cosD = 1/2
   D = 60
因此 ∠ADB = 180-D = 120