Question

show that the figure formed by joining the midpoints of sides of a rhombus successively is a rectangle

Answer 1

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 sides is parallel and equal)

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

Also, AC ⊥ BD (Given)

∴RS ⊥ QR. 

Thus, PQRS is a rectangle.

Answer 2

Answer:

Step-by-step explanation: