Question

Prove that the length of the tangents drawn from an external point to a circle are equal.

Answer 1

Given:PT and TQ are two tangent drawn from an external point T to the circle C (O,r).
To prove:
1. PT = TQ
2. ∠OTP = ∠OTQ
Construction: Join OT.
Proof: We know that,a tangent to circle is perpendicular to the radius through the point of contact.
∴ ∠OPT = ∠OQT = 90°
In ΔOPT and ΔOQT,
OT = OT (Common)
OP = OQ ( Radius of the circle)
∠OPT = ∠OQT (90°)
∴ΔOPTΔOQT (RHS congruence criterion)
⇒PT = TQ and
∠OTP = ∠OTQ (CPCT)
PT = TQ,
∴The lengths of the tangents drawn from an external point to a circle are equal.∠OTP = ∠OTQ.

Answer 2

$\huge\bf{hey }$

$\huge{ \mathfrak{ \pink{here \: is \: answer}}}$

$\bf{ \green{given}}$

tangent AB and AC are on points B and C.

$\bf{to \: prove}$
AB=AC

$\bf{ \blue{construction}}$
join O to A

O to C

and..

O to B

$\bf{ \purple{proof}$
$\sf{in \triangle \: aoc \: and \: \triangle \: aob}$
$\sf{ \angle \: oca = \angle \: oba( \: both \: 90 \degree)}$
$\sf{ob = oc \: \: (radius \: of \: circle)}$
$\sf{oa = oa \: (common)}$
$\sf{ hence\: \triangle \: aoc \: \cong\: \triangle \: aob(by \: rhs)}$

hence

$\bf{ab = ac \: \: (cpct)}$

$\huge{lhs \: = \: rhs}$
$\bf{ \underline{proved}}$

$\huge \boxed{ \boxed{ \orange{hope \: it \: helps}}}$
$\huge{ \gray{thanks}}$