Question

show that quadrilateral formed by joing the midpoints of the pairs of adjacent sides of a rectangle is a rhombus

Answer 1

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

Answer 2

Given AC,BD are diagonals of a quadrilateral ABCD are perpendicular. P,Q,R and S are the mid points of AB,BC, CD and AD respectively. Proof: In ΔABC, P and Q are mid points of AB and BC respectively. ∴ PQ|| AC and PQ = ½AC ..................(1) (Mid point theorem) Similarly in ΔACD, R and S are mid points of sides CD and AD respectively. ∴ SR||AC and SR = ½AC ...............(2) (Mid point theorem) From (1) and (2), we get PQ||SR and PQ = SR Hence, PQRS is parallelogram ( pair of opposite

Now, RS || AC and QR || BD.

Also, AC ⊥ BD (Given)

∴RS ⊥ QR.

Thus, PQRS is a rectangle.