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.)


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.


Beginning from the left-up vertex, name the trapezium vertices $\;A,B,C,D\;$, with the shortest basis up $\;AB\;$, the largets one down $\;CD\;$ , and define


Now use the usual dot product $\;\langle\,,\,\rangle\;$ , and (do the complete claculations):

$$||AC||^2=||BD||^2\iff \langle u+v\,,\,u+v\rangle\iff(t^2-1)||u||^2-2(t+1)\langle u,v\rangle=0\stackrel{\text{(why?)}}\iff$$

$$\iff t-1=\frac{2\langle u,v\rangle}{||u||^2}$$

and now, upon substituting corresponding stuff:



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