Q5 of 62 Page 193

If all sides of a parallelogram touch a circle, show that the parallelogram is a rhombus.

Lengths of tangents drawn from an exterior point to the circle are equal.


From the above theorem,


AP = AS


DS = DR


CR = CQ


BQ = BP


Adding above equations,


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


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


⇒ AB + CD = AD + BC


Also,


AB = CD


AD = BC


[∵ Parallel sides of a parallelogram are equal]


⇒ AB + CD = AD + BC


⇒ AB + AB = BC + BC


⇒ 2AB = 2BC


⇒ AB = BC


⇒ AB = BC = CD = AD


⇒ ABCD is a rhombus


Hence, Proved


More from this chapter

All 62 →
3

AB and CD are two chords of a circle which intersect each other externally at P

(i) If AB = 4 cm BP = 5 cm and PD = 3 cm, then find CD.


(ii) If BP = 3 cm, CP = 6 cm and CD = 2 cm, then find AB.

 4

A circle touches the side BC of ΔABC at P, AB and AC produced at Q and R respectively, prove that AQ = AR =( perimeter of ΔABC)

 6

A lotus is 20 cm above the water surface in a pond and its stem is partly below the water surface. As the wind blew, the stem is pushed aside so that the lotus touched the water 40 cm away from the original position of the stem. How much of the stem was below the water surface originally?

 7

A point O in the interior of a rectangle ABCD is joined to each of the vertices A, B, C and D. Prove that OA2 + OC2 = OB2 + OD2