Q. 114.4( 426 Votes )

# Prove that the parallelogram circumscribing a circle is a rhombus.

Answer :

Since ABCD is a parallelogram,

AB = CD ...(1)

BC = AD ...(2)

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.

Rate this question :

How useful is this solution?

We strive to provide quality solutions. Please rate us to serve you better.

PREVIOUSProve that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.NEXTA triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.

Related Videos

Quiz | Imp. Qs. on Circles37 mins

Short Cut Trick to Find Area of Triangle43 mins

Quiz | Area Related with Circles47 mins

Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Expertsview all courses

Dedicated counsellor for each student

24X7 Doubt Resolution

Daily Report Card

Detailed Performance Evaluation