A quadrilateral ABCD is drawn to circumscribe a circle. Prove that
AB + CD = AD + BC.
∴ AP = AS ...(i)
BP = BQ ...(ii)
CR = CQ ...(iii)
and DR = DS ...(iv)
Adding (i), (ii), (iii) and (iv), we get
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR)
= (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
Hence, AB + CD = BC + DA.
Prove 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.
A 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.
Fig, 10.14
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.