Prove that a parallelogram circumscribing a circle is a rhombus.

Given: ABCD is a ||gm.

To Prove: ABCD is a rhombus.

Theorem Used:

The length of two tangents drawn from an external point are equal.

Explanation:

Let ABCD is a ||gm and its sides touches the circle with centre O.

As A is external point and AS and AP are tangents,

By theorem stated above,

AP = AS …. (1)

Similarly, for point B,

BP = BQ …. (2)

For point C,

CR = CQ …. (3)

For point D,

DR = DS …. (4)

Add 1,2,3 and 4 to get

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

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

AB + CD = AD + BC

As ABCD is a ||gm,

AB = CD

AB = AD = BC = CD

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Quiz | Imp. Qs. on Circle35 mins
Quiz | Testing Your Knowledge on Circles32 mins
Quiz | Imp. Qs. on Circles37 mins
Short Cut Trick to Find Area of Triangle43 mins
Quiz | Areas Related to Circles43 mins
RD Sharma | Area of Sector and Segments25 mins
Quiz | Area Related with Circles47 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses