Mathematics
Answer
As, from point A, AP and AR are the tangents to the circle.
We know that,
If two tangents are drawn to a circle from an exterior point, the tangents are equal in length.
So, we have AP = AR ……….(1)
From point B, BP and BQ are the tangents to the circle.
∴ BP = BQ ……….(2)
From point C, CQ and CR are the tangents to the circle.
∴ CQ = CR …………(3)
Adding equations (1), (2) and (3), we get :
⇒ AP + BP + CQ = AR + BQ + CR
⇒ (AP + BP) + CQ = (AR + CR) + BQ
⇒ AB + CQ = AC + BQ
Given,
AB = AC
∴ BQ = CQ.
Hence, proved that BQ = CQ.
Related Questions
If the sides of a parallelogram touch a circle, prove that the parallelogram is a rhombus.
From a point P outside the circle, with centre O, tangents PA and PB are drawn. Prove that:
(i) ∠AOP = ∠BOP
(ii) OP is the ⊥ bisector of chord AB.
Radii of two circles are 6.3 cm and 3.6 cm. State the distance between their centers if :
(i) they touch each other externally,
(ii) they touch each other internally.