Question

If from an external point T, two tangents TP and TQ are drawn to a circle with centre O. Prove that PQ and OT are right bisectors of each other given that PO is perpendicular to QO

Answer 1

Step-by-step explanation:

Given:-

from an external point T, two tangents TP and TQ are drawn to a circle with centre O.

To find:-

Prove that PQ and OT are right bisectors of each other .

Solution:-

Given that

'O' is the centre of the given circle.

'T' is the external point from the centre of the circle 'O'.

PT and PQ are the two tangents .

Now ,

Construct a straight line from P to Q .

Join O and T and A is the point on OT

∆PTA and ∆QTA are the two triangles

PT = PQ

(The tangents drawn from the external point are equal in length )

angle TPA = angle TQA (opposite angles of equal sides)

TA = TA (Common side)

∆PTA = ∆QTA by SAS Property

now,

PA = AQ (Congruent parts in Congruent triangle are equal )

and angle PAT = angle QAT ------(1)

But

angle PAT + angle QAT = 180°

=>angle PAT + angle PAT = 180°

=>2(angle PAT) = 180°

=>angle PAT = 180°/2

=>angle PAT = 90°

TA is perpendicular to PQ

=>OT is perpendicular to PQ

=>TA or OT is the right bisector of linesegment PQ

Hence, Proved

Used formulae:-

• The tangents drawn from the external point are equal in length.
• The two sides and included angle of a triangle are equal to the two sides and the included angle then the two triangles are Congruent and this property is known as Side-Angle-Side (SAS) property.