Show that the quadrilateral formed by joining the mid pts of the consecutive sides of a rectangle is a rhombus.
MID PTS THEOREM
let ABCD is rectangle.
E,F,G,H be the mid points of the sides AB,BC,CD,DA resply.
EH II BD
EH=1/2[BD].... mid pt thm.
EH II GF
EFGH is a parallelogram... one pair of opp sides is parallel and equal.
by the same argument as above,
we see that
BD=AC...Diagonals of a rectangle are equal.
so we see that for the parallelogram EFGH, all sides ae equal, so it's a rhombus.