Question

Prove that tangents drawn from an external point to a circle are equal in length

Answer 1

construction=join OP,OQ and,OA
proof=in orderAP=AQ,FIRST WE will prove that triangle OPQ congruent triangle OQA
sincea tangent at any point of  a circle is perpendicular to the radius through the point of contact
so OP PERPENDICULAR AP and OQ perpendicular AQ
now=angle OPA=angle OQA=90 degree........(1)
in right angle triangle
OP=OQ......(radiii of circle)
angle OPA= angle OQA(from 1)
and ,OA=OA (common)
so by rhs-criterion of congruency
triangle OPA congruent triangle OQA
SO AP=AQ

Answer 2

Given:-
PQ and TQ are two tangent drawn from an external point T to the circle C
To prove:-
PT=TQ
Angle OTP=OTQ
Construction join OT
Proof:-
We know that, a tangent to circle is perpendicular to the radius through the point of contact.
Since, Angle OPT=OQT=90°
In ∆OPT and ∆OQT,
OT=OT (common side)
OP=PQ (radius of the circle)
Angle OPT=OQT (90°)
Since, OPT≈OQT (RHS)
PT=TQ and Angle OTP=OTQ (CPCT)
PT=TQ
Since, the lengths of the tangent drawn from an external point to a circle are equal.
Angle OTP=OTQ,
Since, centre lies on the bisector of the angle between the two tangent.
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