AD=BC [ Given ]
The mid-point theorem states that the segment joining two sides of triangle at the mid-points of those sides is parallel to the third side and is half the length of the third side.
P and S are the mid-points of sides AB and BD.
So, By mid-point theorem,
⇒ PS∥AD and PS=12AD ------- ( 1 )
R and Q are the mid-point of CD and AC.
So, by mid-point theorem,
⇒ OR∥AD and QR=12AD ----- ( 2 )
Compare ( 1 ) and ( 2 ), we get
⇒ PS∥QR and PS=QR
Since, one pair of opposite sides is equal as well as parallel then,
⇒ PQRS is a parallelogram ---- ( 3 )
Now, In △ABC, by mid-point theorem
⇒ PQ∥BC and PQ=12BC ----- ( 4 )
And AD=BC ----- ( 5 )
Compare equations ( 1 ), ( 4 ) and (5) we get,
⇒ PS=PQ ----- ( 6 )
Since, PQRS is a parallelogram with PS=PQ then PQRS is a rhombus.