In a trapezium ABCD, AD||BC and angle ABC= angle BCD. Let's prove that ABCD is an isosceles trapezium.

$AB$ is the hypotenuse of right isosceles triangle $ABC$. $AD$ is the bisector of angle $\angle BAC$ and $AD$ intersects $BC$ at $D$. Let's prove that $AC+CD=AB$

Two isosceles triangles $ABC$ and $DBC$ whose $AB=AC, DB=DC$ and they are situated on the opposite side of $BC$. Let's prove that $AD$ bisects $BC$ perpendicularly.

Two line segments $PQ$ and $RS$ intersect each other at $X$ in such a way that $XP=XR$ and $\angle PSX=\angle RQX$. Let's prove that $△PXS\cong △RQX$.