Question

TP and TQ are tangents from T to a circle with centre O. At R, a tangent is drawn meeting PT at A and QT at B. Prove that AB=AP+BQ.

Answer 1

Hence proved
This answer is proved by property 'external tangents from a point outside a circle are equal'

Answer 2

Tangents of the circle = TP and TQ (Given)

Centre of the circle = O  (Given)

The tangents drawn from an external point to the circle are always equal in length.

Let T be the external point which is equal in length thus,

TP = TQ    

= TA + AP = TB + BQ --- eq 1

Let A be the external point which is also equal in length, thus,

AP = AR --- eq 2

Let B be the external point, thus,

BQ = BR --- eq 3

Substituting the value of AP and BQ from equation 1, 2, and 3, -

TA + AR = TB + BR

Hence proved.

don't forget to follow..... and rate it if u like ....