Example 2 : Two tangents TP and TQ are drawn
to a circle with centre O from an external point T.
Prove that Z PTQ = 2 Z OPQ
Can you explain this
question briefly.
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We know that length of taughts drawn from an external point to a circle are equal
∴ TP=TQ−−−(1)
4∴ ∠TQP=∠TPQ (angles of equal sides are equal)−−−(2)
Now, PT is tangent and OP is radius.
∴ OP⊥TP (Tangent at any point pf circle is perpendicular to the radius through point of cant act)
∴ ∠OPT=90
o
or, ∠OPQ+∠TPQ=90
o
or, ∠TPQ=90
o
−∠OPQ−−−(3)
In △PTQ
∠TPQ+∠PQT+∠QTP=180
o
(∴ Sum of angles triangle is 180
o
)
or, 90
o
−∠OPQ+∠TPQ+∠QTP=180
o
or, 2(90
o
−∠OPQ)+∠QTP=180
o
[from (2) and (3)]
or, 180
o
−2∠OPQ+∠PTQ=180
o
∴ 2∠OPQ=∠PTQ−−−− proved
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