Q5 of 31 Page 218

Let us prove that a parallelogram circumscribed by a circle is a rhombus.



Let ABCD be the parallelogram


AP and AS are tangents drawn from point A


BP and BQ are tangents drawn from point B


CR and CQ are tangents drawn from point C


DR and DS are tangents drawn from point D


Each of the above pairs of tangents are equal to each other in length since tangents drawn from an external point are always equal.


Using this concept we can say


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


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


Adding each length segment we can say


⇒ AB + CD = AD + BC


Since Opposite sides of a parallelogram are equal so


2AB = 2AD


⇒ AB = AD


Since adjacent sides of a parallelogram are equal, so it’s a rhombus.


More from this chapter

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3

AP and AQ are two tangents drawn from an external point A to a circle with centre O, P and Q are points of contact. If PR is a diameter, let us prove that OA ∥ RQ

 4

Let us prove that for a quadrilateral circumscribed about a circle, the angles subtended by any two opposite sides at the centre are supplementary to each other.

 6

Two circle drawn with centre A and B touch each other externally at C, O is a point on the tangent drawn at C, OD and OE are tangents drawn to the two circles of centers A and B respectively. If ∠ COD = 56°, ∠COE = 40, ∠ACD = x° and ∠BCE = y°. Let us prove that OD = OC = OE and y-x = 8

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Two circles with centers A and B touch each other internally. Another circle touches the larger circle externally at the point x and the smaller circle externally at the point y. If O be the centre of that circle, let us prove that AO + BO is constant.