MID PTS THEOREM

Asked by  | 13th Sep, 2009, 06:46: AM

Expert Answer:

let ABCD is  rectangle.

E,F,G,H  be the mid points of the sides AB,BC,CD,DA resply.

join BD.

EH  II BD

and

 EH=1/2[BD].... mid pt thm.

simly

GFII BD

and

 GF=1/2[BD]

So

EH=GF

and

EH II GF

so

EFGH is a parallelogram... one pair of opp sides is parallel and equal.

by the same argument as above,

 we see that

EF=GH=1/2[AC]

but

 BD=AC...Diagonals of a rectangle are equal.

so

EF=GH=1/2[AC]=1/2[BD]=EH=GF

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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