From an external point P, two tangents, PA and PB are drawn to a circle with centre O. At one point E on the circle tangent is drawn which intersects PA and PB at C and D, respectively. If PA = 10 cm, find the perimeter of the triangle PCD.
In this figure
CE = CA [ Tangents from an external point to a circle are equal in length]
Similarly,
DE = DB and PB = PA
Perimeter of DPCD
= PC + CD + PD
= PC + CE + ED + PD ( CD = CE + ED)
= PC + CA + DB + PD
= PA + PB [ PC + CA = PA and DB + PD = PB]
= PA + PA [ PB = PA]
= 2PA
= 2 × 10 [ PA = 10]
= 20 cm