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

Answer is:-

Tangents of the circle = TP and TQ (Given)

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.

Hope it helps you dear.......xd

Answer 2

Answer:

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