Question

From a point T outside a circle of centre O, tangents TP and TQ are drawn to the circle. Prove that OT is the right bisector of line segment PQ.

Answer 1

we know : tangent at any point of a circle is perpendicular to radius.
so <opt=< oqt= 90degree
In triangle opt and triangle oqt
OP=OQ ( radii of same circle)
<OPT=<OQT =90degree
OT=OT ( common)
so triangle OPT=OQT
so PT = PQ ( by c.p.ct)
so in quadrilateral OPTQ
OP= OQ; PT=QT and <OPT=<OQT
so it is a kite
nd v know in kite diagonals are perpendicular to each other and longer diagonal bisect smaller diagonal.

Answer 2

Given : A circle with centre O. Tangents TP  and TQ are drawn from a point T outside a  circle.

To Prove : OT is the right bisector of line  segment PQ.

Construction : Join OP & OQ

Proof : In ΔPTR and ΔQTR

In ΔOPT and ΔOQT

∠OPT = ∠OQT = 90°

OP = OQ (radius)

OT = OT (Common)

ΔOPT ≅  ΔOQT (By RHS congruence)

∠PTR = ∠QTR (cpct)

TP = TQ  (Tangents are equal)

TR = TR (Common)

∠PTR = ∠QTR  (OT bisects ∠PTQ)

ΔPTR ≅ ΔQTR (By SAS congruency)

PR = QR

∠PRT = ∠QRT

But ∠PRT+ ∠QRT = 180° (as PQ is line segment)

∠PRT = ∠QRT = 90°

Therefore TR or OT is the right bisector of  line segment PQ