Question

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

Answer 1

Step-by-step explanation:

Hello Mate!

Given : ABCD is rhombus where PQRS are formed by joining mid points.

To prove : PQRS is a rectangle.

To construct : Join AC and BD.

Proof : Since RQ are formed by joining mid points so, QR || BD and QR = ½ BD _(i)

Since PS are formed by joining mid points so, PS || BD and PS = ½ BD _(ii)

From (i) and (ii) we get,

QR || PS and QR = PS

Hence a pair if side is equal and parallel so PQRS is ||gm.

Now, since QR || BD so MR || ON __(iii)

Similarly, NR || AC __(iv) [ Because SR || AC, since SR is formed by joining mid points ]

From (iii) and (iv) we get, MONR is a ||gm.

As we know that diagonals of rhombus bisect at 90° so < MON = 90°

Now, since MONR is ||gm so its opposite angles will be equal. So,

< MRN = 90°

Or < QRS = 90°

Hence PQRS is such ||gm whose one if the angle ( < QRS ) is 90° so PQRS is a rectangle.

ʜᴇɴᴄᴇ ᴘʀᴏᴠᴇᴅ.