Prove that the parallelogram inscribed a circle is rhombus.
Answers
Given: ABCD be a parallelogram circumscribing a circle with centre O.
To prove: ABCD is a rhombus.
We know that the tangents drawn to a circle from an exterior point are equal in length.
Therefore, AP = AS, BP = BQ, CR = CQ and DR = DS.
Adding the above equations,
AP + BP + CR + DR = AS + BQ + CQ + DS
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC
2AB = 2BC
(Since, ABCD is a parallelogram so AB = DC and AD = BC)
AB = BC
Therefore, AB = BC = DC = AD.
Hence, ABCD is a rhombus.
Let ABCD is a ||gm which inscribed a Circle.
To Prove :- AB = BC.
Proof :- AB + CD = AD + BC ( Opposite sides of ||gm)
Taking L.H.S,
AB + CD = AE + BE + DG + CG
AB +CD = AH + BF + CF + DH
AB + CD = AH + DH + BF + CF
AB + CD = AD + BC
AB + AB = BC + BC
2 AB = 2 BC
AB = BC
Proved