Question

show that the quadrilateral formed by joining the mid points of a rectangle in order is a rhombus​

Answer 1

Step-by-step explanation:

Let the rhombus have vertices A, B, C, D. Let the midpoints of the sides AB, BC, CD, DA be E, F, G, H. If the diagonals of EFGH are equal and bisect each other then EFGH is a rectangle

Answer 2

Let the rhombus have vertices A, B, C, D. Let the midpoints of the sides AB, BC, CD, DA be E, F, G, H.

If the diagonals of EFGH are equal and bisect each other then EFGH is a rectangle.

AB || HF || DC and AD || EG || BC.

So HF = AB = BC = EG.

ie diagonals HF and EG are equal in length.

HF bisects AD and BC. EG is parallel to AD and BC. Therefore HF bisects EG also.

By a similar argument, EG bisects HF.

So diagonals HE and EG (i) are equal in length, and (ii) bisect each other.

Therefore EFGH is a rectangle