Question

Prove that the figure formed by joining the mid points of the adjacent sides of a rectangle is a rhombus.

Answer 1

Given: ABCD is a rectangle.
P, Q, R, and S is the mid-point of AB, AD, DC, and CB, respectively.
To prove: $PQRS$ is a rhombus.
Proof:
Joint $PQ,QR,RS,ST$ (constructions)
$PQ=RS=0.5AC$
$PS=QR=0.5BD$ (Mid-point theorem)
$\Rightarrow$ $PQRS$ is a parallelogram ( The opposite sides of a quadrilateral areparallel and equal)
$AC=BD$ (the diagonals of a rectangle equal)
$\Rightarrow$ $PQ=RS=QR=PS$ (the half of the equal quantities equal)
$PQRS$ is a rhombus (4 sides are equal).