Question

Prove that if the two angles on the base of a trapezoid are different, then the diagonal starting from the smaller angle is longer than the other diagonal.

Answer 1

$\angle ABC<\angle DCB$ iff A'B > CD'
$putA{A}^{\prime }=h,A^{\prime }{D}^{\prime }=a,thenB{D}^2=h^2+(a+A^{\prime }B{)}^2$
$A{C}^2=h^2+(a+C{D}^{\prime }{)}^2$
So $BD>AC$ iff $A^{\prime }B>C{D}^{\prime }$