Question

Answer the question ​

Answer 1

Answer:

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Since ABCD is a parallelogram,

AB = CD ---- i)

BC = AD ---- ii) 

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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.

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Answer 2

$\large{\mathscr{\green{\underline{\underline{\:\:\:\:\:\:Given\:\:\:\:\:\:}}}}}$

• $\bf{ABCD\:is\:a\:parallelogram\:circumscribing\:a\:circle}$

$\large{\mathscr{\green{\underline{\underline{\:\:\:\:\:\:To\:\:prove\:\:\:\:\:\:}}}}}$

• $\bf{ABCD\:is\:a\:rhombus}$

$\large{\mathscr{\green{\underline{\underline{\:\:\:\:\:\:Proof:-\:\:\:\:\:\:}}}}}$

$\sf{Since,the\:tangents\:from\:an\:external\:point\:to\:a\:circle\:are\:equal}$

• $\sf{So,}$

$\tt{AP=AS}$_______(i)

$\tt{BP=BQ}$_______(ii)

$\tt{CR=CQ}$_______(iii)

$\tt{DR=DS}$_______(iv)

$\sf{\underline{On\:adding\:eq.(i),(ii),(iii)\:and\:(iv), we\:get}}$

$\tt{(AP+BP)+(CR+DR)=(AS+BQ)+(CQ+DS)}$

$\tt{⇒ AB+CD=(AS+DS)+(BQ+CQ)}$

$\tt{⇒ AB+CD=AD+BC}$

$\tt{⇒ AB+AB=AD+AD}$

⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\bf{(Opposite\:sides\:of\:||\:gm\:are\:equal)}$

$\tt{⇒ 2AB=2AD}$

$\tt{⇒AB=AD}$

$\sf{But\:AB=CD\:and\:AD=BC}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\bf{(Opposite\:sides\:of\:||\:gm)}$

$\sf{So,AB=BC=CD=AC}$

$\bf{So,\:parallelogram\:ABCD\:is\:a\:rhombus}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\large{\mathscr{\green{\underline{\:\:\:\:\:\:Hence\:proved\:\:\:\:\:\:}}}}$