# prove that the parallelogram circumscribing a circle is a rhombus.???

Since ABCD is a parallelogram,

AB = CD …(1) It can be observed that

DR = DS (Tangents on the circle from point D)

CR = CQ (Tangents on the circle from point C)

BP = BQ (Tangents on the circle from point B)

AP = AS (Tangents on the circle from point A)

Adding all these equations, we obtain

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

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

CD + AB = AD + BC

On putting the values of equations (1) and (2) in this equation, we obtain

2AB = 2BC

AB = BC …(3)

Comparing equations (1), (2), and (3), we obtain

AB = BC = CD = DA

Hence, ABCD is a rhombus.

• 85

But i would like to point something....

we generally get confused in adding up all the equations.. so as to get, CD + AB = AD + BC.

so do this

given, AB = CD and AD = BC

now consider,

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

=> AD + BC = AB + CD

thus, AB = BC = CD = AD

Thus, ABCD is a parallelogram in which all the sides are equal...

and hence, ABCD is a rhombus... njoy

• 13

Since ABCD is a parallelogram,

AB = CD …(1) It can be observed that

DR = DS (Tangents on the circle from point D)

CR = CQ (Tangents on the circle from point C)

BP = BQ (Tangents on the circle from point B)

AP = AS (Tangents on the circle from point A)

Adding all these equations, we obtain

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

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

CD + AB = AD + BC

On putting the values of equations (1) and (2) in this equation, we obtain

2AB = 2BC

AB = BC …(3)

Comparing equations (1), (2), and (3), we obtain

AB = BC = CD = DA

Hence, ABCD is a rhombus.

• 7

thank you anna

• 3

THE SIMPLEST METHOD BELOW
Since ABCD is a parallelogram,

AB = CD …(1) It can be observed that

DR = DS (Tangents on the circle from point D) = x

CR = CQ (Tangents on the circle from point C) = y

BP = BQ (Tangents on the circle from point B) = z

AP = AS (Tangents on the circle from point A) = w

Adding all these equations, we obtain

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

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

CD + AB = AD + BC

On putting the values of equations (1) and (2) in this equation, we obtain

2AB = 2BC

AB = BC …(3)

Comparing equations (1), (2), and (3), we obtain

AB = BC = CD = DA

Hence, ABCD is a rhombus.

• 3

the easiest method ...!!

BELOW
Since ABCD is a parallelogram,

AB = CD …(1) It can be observed that

DR = DS (Tangents on the circle from point D) = x

CR = CQ (Tangents on the circle from point C) = y

BP = BQ (Tangents on the circle from point B) = z

AP = AS (Tangents on the circle from point A) = w

Adding all these equations, we obtain

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

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

CD + AB = AD + BC

On putting the values of equations (1) and (2) in this equation, we obtain

2AB = 2BC

AB = BC …(3)

Comparing equations (1), (2), and (3), we obtain

AB = BC = CD = DA

Hence, ABCD is a rhombus.

• 3

why is the sky so high?

• 2

Figure ½ m

ABCDis a parallelogramcircumscribing a circlewith centre OAP =AS, DS = DR, CR = CQ, PB = BQ 1 m (AP + PB) + (DR + CR) = (AS + DS) + (BQ + QC) ½ mÞ AB + DC =AD + BC or 2AB = 2AD ½ mÞ AB =AD A BCD is a rhombus ½ m
• 1
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