Question

Prove that the length of tangent drawn from the external point to the circle are equal

Answer 1

Hey there!

→ Given: PT and TQ are two tangent drawn from an external point T to the circle C (O, r).

→ To prove: PT=TQ ; Tangent drawn from the external point to the circle are equal. 


→ 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°)

∴ ΔOPT ΔOQT  (RHS Congruence criterion)

⇒ PT = TQ  
∴ The lengths of the tangents drawn from an external point to a circle are equal.

HOPE IT HELPED ^_^

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

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

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

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