Question

in the given figure PO is perpendicular to QO. the tangent to the circle with Centre O at P and Q intersect at point a prove that PQ and OT are right bisector of each other

Answer 1

angle opt=angle oqt=90 degree
angle t =90 degree(angle sum property)
PTQO is a square(llgm with all angles 90 degree and adjacent sides are equal)
in a square the diagonals are ⊥ bisector 
∴pq and ot are perpendiculaer bisector 

Answer 2

Hence proved that PQ and OT are right bisectors of each other.

Step-by-step explanation:

Given that,

PO⊥ QO while The tangents P and Q of the circle intersect at the point T.

In ΔTPO and ΔTQO,

TQ = PT      (∵ Tangents drawn from the external point are equal in length)

OT = OT     (Common)

∠TPO =∠TQO = 90°

Thus, by using the RHS rule, we have

ΔTPO ≅ ΔTQO

so, ∠PTO = ∠QTO     ...(1)     [using C.P.C.T.]

In ΔPTR and ΔQTR

PT = TQ [∵Tangents Tangents drawn from an external point are equal in length)

∠PTO = ∠QTO                [by (1)]

TR = TR [Common]

So, using SAS rule;

ΔPTR ≅ ΔQTR

∴ PR = RQ             ...(2)

And,

∠TRP = ∠TRQ

Though, ∠TRP + ∠TRQ = 180∘

⇒ 2∠TRP =180∘

⇒∠TRP = 90∘        ...(3)

Hence, it's proved that PQ and OT are right bisectors of one another.

Learn more: Perpendicular Bisector

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