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Let as in the question , ABCD be the trapizium with AD||BC ; w.r.t. some chosen origin let
 AD=x⃗  and DC=y⃗  , then CB=−tx⃗  for some t>0 . Now
AB=AD+DC+CB=x⃗ +y⃗ −tx⃗  , AC=AD+DC=x⃗ +y⃗  and
DB=DC+CB=y⃗ −tx⃗ . Now given |AC|=|DB| , so
0=|AC|2−|DB|2=(ACDB).(AC+DB)=(x⃗ +tx⃗ ).(x⃗ +2y⃗ −tx⃗ )
=((1+t)x⃗ ).(x⃗ +2y⃗ −tx⃗ )=(1+t)(x⃗ .(x⃗ +2y⃗ −tx⃗ )) [using m(a⃗ .b⃗ )=(ma⃗ ).b⃗ ]
hence either 1+t=0 , or x⃗ .(x⃗ +2y⃗ −tx⃗ )=0 , now in the former case t=−1<0 ,contradiction! hence x⃗ .(x⃗ +2y⃗ −tx⃗ )=0 ....(i) . Hence
|AB|2−|DC|2=(ABDC).(AB+DC)=(x⃗ −tx⃗ ).(x⃗ +2y⃗ −tx⃗ )
=((1−t)x⃗ ).(x⃗ +2y⃗ −tx⃗ )=(1−t)(x⃗ .(x⃗ +2y⃗ −tx⃗ ))=0  [using (i)] , and this was
to be proved.
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