Two tangents TP and TQ are drawn to a circle with center O from an external point T. Prove that angle PTQ = 2 OPQ.

We know that, the lengths of tangents drawn from an external point to a circle are equal.

TP = TQ

In ΔTPQ,

TP = TQ

TQP = TPQ ...(1) (In a triangle, equal sides have equal angles opposite to them)

TQP + TPQ + PTQ = 180º (Angle sum property)

2 TPQ + PTQ = 180º (Using(1))

PTQ = 180º – 2 TPQ ...(1)

We know that, a tangent to a circle is perpendicular to the radius through the point of contact.

OP PT,

OPT = 90º

OPQ + TPQ = 90º

OPQ = 90º – TPQ

2OPQ = 2(90º – TPQ) = 180º – 2 TPQ ...(2)

From (1) and (2), we get

PTQ = 2OPQ

 

  • 360
What are you looking for?