The roots of the equation(a2+b2)t2-2(ac+bd)t+(c2+d2)=0 are equal, then

Accordingly, {2(ac+bd)}2=4(a2+b2)(c2+d2)⇒4a2c2+4b2d2+8abcd=4a2c2+4a2d2+4b2c2+4b2d2⇒4a2d2+4b2c2-8abcd=0⇒4(ad-bc)2=0  ⇒ad=bc⇒ab=cd.