Question

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

Answer 1

Hey !!

Here is your answer...

↪Given :- P is a point in the exterior of circle.
Tangent to the circle from P touches the circle at T1 and T2 .

↪To prove :- PT1 = PT2

↪Proof :- Join O to P.

In ∆ OPT1 and ∆ OPT2, consider the correspondence OPT1 ↔ OPT2.

angle OT1P = angle OT2. .........( right angles )
OT1 = OT2 .........( radius )
OP = OP ..........( common )

∆ OPT1 = ∆ OPT2. ........( by RHS Theorem. )
Hence, PT1 = PT2

PT1 = PT2 .......( proved )


Hope it helps you....

Thanks..
^-^

Answer 2

$\huge \bf{ \red{hey}}$

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

$\bf{ \green{given}}$

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

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

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

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