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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» 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 |
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