Question

show that quadrilateral formed by the joining the midpoints of the sides of a rhombus taken in order, forms an rectangle.​
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Answer 1

Answer:

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.

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