Q:

To prove by vector method , non-parallel sides of a trapezium having equal diagonals are equal.

How do we prove by vector method that "if the diagonals of a trapezium have equal length then the non-parallel sides of the trapezium have equal length." ? 
(taking $ABCD$ to be the trapezium with 
$AD$ || $BC$ and $O$ the intersection point of diagonals , if it can be shown by vector method that 
$OB=OC$ , then $AB=CD$ follows ; I can show $OB=OC$ but not by vector method and thus the problem , though any other line of approach is acceptable.)  

A:

Let as in the question , $ABCD$ be the trapizium with $AD||BC$ ; w.r.t. some chosen origin let $\vec {AD}=\vec{x}$ and $\vec {DC}=\vec{y}$ , then $\vec{CB}=-t\vec x$ for some $t>0$ . Now 
$\vec {AB}=\vec{AD}+\vec{DC}+\vec{CB}=\vec x+\vec y-t\vec x$ , $\vec{AC}=\vec{AD}+\vec{DC}=\vec x+\vec y$ and 
$\vec{DB}=\vec{DC}+\vec{CB}=\vec y-t\vec x $. Now given $|\vec{AC}|=|\vec{DB}|$ , so 
$0=|\vec{AC}|^2-|\vec{DB}|^2=(\vec{AC}-\vec{DB}).(\vec{AC}+\vec{DB})=(\vec x+t\vec x).(\vec x+2\vec y -t\vec x)$ 
$=\Big((1+t)\vec x \Big).(\vec x+2\vec y -t\vec x)=(1+t)\Big(\vec x.(\vec x+2\vec y -t\vec x)\Big)$ [using $m(\vec a.\vec b)=(m\vec a).\vec b$] 
hence either $1+t=0$ , or $\vec x.(\vec x+2\vec y -t\vec x)=0$ , now in the former case $t=-1<0$ , 
contradiction! hence $\vec x.(\vec x+2\vec y -t\vec x)=0$ ....(i) . Hence 
$|\vec{AB}|^2-|\vec{DC}|^2=(\vec {AB}-\vec{DC}).(\vec{AB}+\vec{DC})=(\vec x-t\vec x).(\vec x+2\vec y -t\vec x)$
$=\Big((1-t)\vec x \Big).(\vec x+2\vec y -t\vec x)=(1-t)\Big(\vec x.(\vec x+2\vec y -t\vec x)\Big)=0$$\space$ [using (i)] , and this was 
to be proved. 

