Question

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

Answer 1

Heya !!

Here is your answer..

Theorem :

The lengths of the tangents drawn from an external point to a circle are equal.

Let C be a circle with centre O and P be a external point.

Let PT and PS be the two tangents from the point P to the circle C.

S and T be the point of contacts.

Construction :

Join PO , OS and OT.

Proof :

In triangle PTO & PSO ,

OT = OS {radii}

angle PTO = angle PSO {each 90° since Tangent is perpendicular to radius}

PO = PO {common}

==> Triangle PTO congruent to PSO {RHS}

==> PT = PS { by CPCT }

==> Lengths of tangents drawn from an external point are equal.

PROVED.

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

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

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

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