Question

Prove that the parallelogram circumscibing a circle is rhombus​

Answer 1

Given : ABCD is a parallelogram circumscribing a circle with centre O and let P, Q, R and S be the point on the parallelogram that touch the circle.

To prove : ABCD is a rhombus.

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Proof : In a parallelogram ABCD :

• AB = CD (Equal) . . . . . (1)
• AD = BC (Equal) . . . . . (2)

Also, we know that, tangents to a circle from same point are equal in length. So :

• AP = AS . . . . . (3)
• DR = DS . . . . . (4)
• CR = CQ . . . . . (5)
• BP = BQ . . . . . (6)

Adding (3), (4), (5) and (6) :

• AP + DR + CR + BP = AS + DS + CQ + BQ
• (AP + BP) + (DR + CR) = (AS + DS) + (CQ + BQ)
• AB + CD = AD + BC

Substituting the values of (1) and (2) in the above equation :

• AB + CD = AD + BC
• AB + AB = BC + BC
• 2AB = 2BC
• AB = 2BC/2
• AB = BC . . . . . . (3)

Thus, from (1), (2) and (3), we get :

• AB = BC = CA = AD

We know that in a rhombus, all sides are equal.

Therefore, ABCD is a rhombus.

Henceforth, proved!

Answer 2

Step-by-step explanation:

Let ABCD be a parallelogram with sides AB,BC,CD and DA and they touch a circle at the points P,Q,R,S respectively.

Since the lengths of tangents are drawn from a external point to a circle are equal, we have:

AB = AS, BP = BQ, CR = CQ, DR = DS

∴ AB + CD = AP + BP + CR + DR

i.e.,

AS + BQ + CQ + DS

(AS + DS) + (BQ + CQ)

AD + BC

Now,

AB + CD = AD + BC

2AB = 2BC (Opp. sides of parallelogram are equal)

∴ AB = BC

∴ AB = BC = CD = AD

Hence, ABCD is a rhombus. (All sides are equal in rhombus)