Since the length of two tangents drawn from an external point are equal.
Therefore, BP = BQ
CP = CR
and AQ = AR
Perimeter of ΔABC,
= AB + BC + AC
= AB + BP + PC + AC
= (AB + BQ) + (CR + AC)
[∵ BP = BQ, CP = CR]
= AQ + AR [ ∵ AQ = AR]
= 2 AQ
It is given that : AQ = 5 cm
Therefore, perimeter (ΔABC) = 2 x 5 = 10 cm.
From a point P outside a circle, with centre O tangents PA and PB are drawn as shown in the figure, prove that
(i) ⇒AOP = ∠BOP
(ii) OP is the perpendicular bisector of AB.