Question

ABCD is a rhombus and P,Q,R and S are the mid points of the sides AB,BC,CD and DA respectively. Show that the quadrilateral PQRS is a rectangle

Answer 1

$\huge\textsf{\underline{\underline{Answer:-}}}$

Given:- In rhombus ABCD

P,Q,R and S are the mid points

of AB=BC=CD=DA.

To prove:- PQRS is a rectangle.

Proof:- In ∆ABD,

P and S are the mid points of

AB And AD as respectively.

Therefore, PS ll BD and PS = $\sf\frac{1}{2}$ (by the mid point theorem)----1.

In ∆CBD

Q and R are the mid points of

BC and CD respectively.

Therefore, QRllBD and QR= $\sf\frac{1}{2}$ BD. (by the mid point theorem)---- 2.

From 1 and 2

Therefore, PS ll QR and PS =QR

Therefore, PQRS IS A RECTANGLE.

Answer 2

Answer:

Given:- In rhombus ABCD

P,Q,R and S are the mid points

of AB=BC=CD=DA.

To prove:- PQRS is a rectangle.

Proof:- In ∆ABD,

P and S are the mid points of

AB And AD as respectively.

Therefore, PS ll BD and PS =  (by the mid point theorem)----1.

In ∆CBD

Q and R are the mid points of

BC and CD respectively.

Therefore, QRllBD and QR=  BD. (by the mid point theorem)---- 2.

From 1 and 2

Therefore, PS ll QR and PS =QR

Therefore, PQRS IS A RECTANGLE.