Question

prove that the parallelogram circum-scribing a circle is a rhombus.​

Answer 1

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➠ Fig. in the above Attachment :-

★$\huge\underline \mathfrak \black {Given:-}$

➠ A circle with the center O .

A parallelogram ABCD touching the circle at points P, Q, R, S.

★$\huge\underline \mathfrak \black {To Prove :-}$

➠ ABCD is a rhombus.

★$\huge\underline \mathfrak \black{Proof :-}$

➠ A rhombus is a parallelogram with all sides equal.

➠ In parallelogram ABCD,

• AB = CD & AD = BC - - - { Opp. Sides of parallelogram are equal }.... (1)

➠ Now, using the theorem of tangents drawn from external points are equal.

➠ Hence,

• AP = AS.... (2)
• BP = BQ.... (3)
• CR = CQ.... (4)
• DR = DS.... (5)

➠ Now adding (2) + (3) + (4) + (5)

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

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

→ AB + CD = AD + BC

➠ From (1) :- CD = AB & BC = AD

→ AB + AB = AD + AD

→ 2 AB = 2 AD

➠ i.e, AB = AD.

➠ So,

→ AB = AD &

→ AB = CD, AD = BC

➠So,

→ AB = CD = AD = CD

→ "So, ABCD is a parellelogram with all sides equal" .

➠ "ABCD is rhombus" .

★ :)_Hence, Proved....!!! ★

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