Question

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

Answer 1

Step-by-step explanation:

Given: A circle with centre O. A parallelogram ABCD touching the circle at points P,Q,R and S To prove: ABCD is a rhombus Proof: A rhombus is a parallelogram with all sides equal, So, we have to prove all sides equal In parallelogram ABCD, AB = CD & AD = BC From theorem 10.2, lengths of tangents drawn from external point are equal Hence, AP = AS BP = BQ CR = CQ DR = DS 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 AB + AB = AD + AD 2AB = 2AD AB = AD So, AB = AD & AB = CD & AD = BC So, AB = CD = AD = CD So, ABCD is a parallelogram with all sides equal ABCD is a rhombus Hence proved

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Answer 2

Step-by-step explanation:

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Answer is there in the pic