如图,AB、CD是圆中互相垂直的直径,在弧BC和弧BD上取点E、F,连接AE、AF,

令圆形为O
设∠OAM = α,∠OAN = β
圆内接四边形对角和为180,因此∠EAF + ∠EBF = 180
∠EBF = 135,因此∠EAF = 45,即α+β=45
设圆半径为x(x>2),则OA = OB = OC = OD = x
因为CM = 2,DN = 1,所以OM = x - 2,ON = x - 1
△AMO中,tanα= OM / OA = (x-2) / x
△ANO中,tanβ= ON / OA = (x-1) / x
1 = tan45 = tan(α+β) = (tanα+ tanβ) / (1 - tanα·tanβ)
即可解得x = (3 + √5) / 2 （舍去了x = (3 - √5) / 2）
即直径AB = (3 + √5)