#### Question

If the sides of a parallelogram touch a circle (refer figure of Q. 7), Prove that the parallelogram is a rhombus

#### Solution

From A, AP and AS are tangents to the circle.

Therefore, AP = AS.......(i)

Similarly, we can prove that:

BP = BQ .........(ii)

CR = CQ .........(iii)

DR = DS .........(iv)

Adding,

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

AB + CD = AD + BC

Hence, AB + CD = AD + BC

But AB = CD and BC = AD.......(v) Opposite sides of a ||gm

Therefore, AB + AB = BC + BC

2AB = 2 BC

AB = BC ........(vi)

From (v) and (vi)

AB = BC = CD = DA

Hence, ABCD is a rhombus

Is there an error in this question or solution?

Solution If the Sides of a Parallelogram Touch a Circle (Refer Figure of Q. 7), Prove that the Parallelogram is a Rhombus Concept: Tangent Properties - If a Line Touches a Circle and from the Point of Contact, a Chord is Drawn, the Angles Between the Tangent and the Chord Are Respectively Equal to the Angles in the Corresponding Alternate Segments.