Question

A circle is under a parallelogram now prove that , that parallelogram is rhombus.

Prove that the parallelogram circumscribing a circle is a rhombus.​

Answer 1

Step-by-step explanation:

Given:

• A parallelogram ABCD circumscribing a circle.

To Prove:

• ABCD is a rhombus?

Solution: See the figure above , we have

• AB = DC {Opposite sires of ||gm}
• AD = BC {Opposite sires of ||gm}

As we know that length of two tangents drawn from a external points to a circle are equal to each other.

Let's see the tangents form each point

• AP = AS {from point A}.....i
• BP = BQ {from B}.....ii
• CR = CQ {from C}.....iii
• DR = DS {from D}.....iv

Now,

➯ AP + BP + CR + DR = BQ + CQ + AS + DS

➯ AB + CD = BC + AD

➯ 2AB = 2AD (opposite sides are equal)

➯ AB = AD

∴ABCD is a ||gm with adjacent sides AB = AD. Hence, is a rhombus.

$\large\bold{\texttt {Proved }}$

Answer 2

Answer:

✡ Given :- ABCD be a parallelogram circumscribing a circle with center O.

✡ To Prove:- ABCD is a rhombus.

We know that the tangents draw to a circle from am exterior point are equal in length.

Therefore, AP = AS, BP = BQ, CR = CQ and DR = DS

Adding the above equation,

$\leadsto$ AP + BP + CR + DR = AS + BQ + CQ + DS

$\leadsto$ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

$\leadsto$ AB + CD = AD + BC

$\leadsto$ 2AB = 2BC

(Since, ABCD is a parallelogram so

AB = DC and AD = BC)

➡ AB = BC

Therefore, AB = BC = DC = AD

Hence, ABCD is a rhombus.

(Proved)

Step-by-step explanation:

HOPE IT HELP YOU