Question

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Answer 1

let the centre of the circle be O and the two tangents forming an angle at point P from points A & B respectively
now, `/_ APB is supplementary to /_AOB `
we have to prove that, `/_APB+ /_AOB = 180^@`
now draw 2 `_|_` OA & OB TO AP & BP respectively
now, `/_OAP = /_OBP= 90^@`
now, in quadilateral AOBP, sum of 4 angles = `360^@`
`/_AOB + /_OBP + /_APB+ /_OAP = 360^@`
as we know that, `/_ OBP = 90^@ & /_OAP=90^@`
so, `/_AOB + /_APB = 180^@`