two tangents TP and tq are drawn to a circle with Centre O from an external point t prove that angle ptq is equals to 2 OPQ
Answers
Answer:
Step-by-step explanation:
Answer:
Step-by-step explanation:
Given a circle with centre O ,
an external point T and two tangents
TP and TQto the circle ,
Where P , Q are the points of contact.
We need to prove that ,
<PTQ = 2<OPQ
Let <PTQ = x°
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By the theorem ,
The lengths of tangents drawn from
an external point to a circle are equal
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TP = TQ ,
So, TPQ is an Isosceles triangle .
Therefore ,
<TPQ + <TQPn+ <PTQ = 180°
[ Angle sum property ]
=> <TPQ = <TQP = ( 180° - x )/2
= 90° - x/2
[ By , theorem
The tangent at any point of a circle is
perpendicular to the radius through
the point of contact ]
<OPT = 90°
<OPQ = <OPT - <TPQ
= 90° - ( 90 - x/2 )
= x/2
= <PTQ/2
This gives ,
<OPQ = <PTQ/2
Therefore ,
<PTQ = 2<OPQ
Similarly
<PTQ = 2<OQP