（a^2+b^2）*d^2-2b(a+c)d+b^2+c^2=0 => a^2*d^2+b^2*d^2-2abd-2bcd+b^2+c^2=0 => (a^2*d^2-2abd+b^2)+(b^2*d^2-2bcd+c^2)=0 => (ad-b)^2+(bd-c)^2=0 => ad-b=0 and bd-c=0 (因为只有每一项都等于0,他们的平方才能等...