Question

Two tangents TP and TQ are drawn to a circle with centre O from an external
point T. Prove that LPTQ = 2LOPQ.
Q​

Answer 1

Answer:

Step-by-step explanation:

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