CBSE Class 9 Answered

Given :- ABCD is a rectangle and E,F,G,H are the midpoints of sides AB, BC< CD and DA respectively.

To Prove:- EFGH is a rhombus.

Proof:-In ABC,

E and F are the mid points of AB and BC respectively.

EF || AC and EF = AC

Similarly, GH || AC and GH = AC

Hence, EF || GH and EF = GH

Thus, EFGH is a parallelogram.

Now, in triangles AEH and EBF,

AE=BE

EAH = EBF

AH = BF

AEH BEF and EH=EF

Therefore, EFGH is a ||^gm in which adjacent sides are equal.

Hence, EFGH is a rhombus