Question

if the diagonals of a trpezium are equal prove that its non parallel sides are equal.

Answer 1

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=(AC→−DB→).(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=(AB→−DC→).(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.