MID PTS THEOREM
Asked by | 13th Sep, 2009, 06:46: AM
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.
Answered by | 13th Sep, 2009, 09:21: AM
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