Prove that the parallelogram circumscribing a circle, is a rhombus.
Prove that the parallelogram circumscribing a circle, is a rhombus.
Given: A parallelogram ABCD circumscribes a circle with center O.
To prove: AB = BC = CD = AD
Proof: we know that the lengths of tangents drawn from an exterior point to a circle are equal.
Therefore, AP = AS .... (i) [tangents from A]
BP = BQ .... (ii) [tangents from B]
CR = CQ .... (iii) [tangents from C]
DR = DS .... (iv) [tangents from D]
Therefore, AB + CD = AP + BP + CR + DR
= AS + BQ + CQ + DS [From (i), (ii), (iii), (iv)]
= (AS + DS) + (BQ + CQ)
= AD + BC
AB = CD, BC = AD (given that ABCD is a parallelogram)
On substituting in the above equation, we get
AB + AB = BC + BC
AB = BC
Therefore, AB = BC = CD = AD
Given: A parallelogram ABCD circumscribes a circle with center O.
To prove: AB = BC = CD = AD
Proof: we know that the lengths of tangents drawn from an exterior point to a circle are equal.
Therefore, AP = AS .... (i) [tangents from A]
BP = BQ .... (ii) [tangents from B]
CR = CQ .... (iii) [tangents from C]
DR = DS .... (iv) [tangents from D]
Therefore, AB + CD = AP + BP + CR + DR
= AS + BQ + CQ + DS [From (i), (ii), (iii), (iv)]
= (AS + DS) + (BQ + CQ)
= AD + BC
AB = CD, BC = AD (given that ABCD is a parallelogram)
On substituting in the above equation, we get
AB + AB = BC + BC
AB = BC
Therefore, AB = BC = CD = AD
