Question

the length of the tangent drawn from an external point to a circle are equal prove​

Answer 1

this is your answer

I hope this answer is right

Answer 2

$\bf \huge \pink{ Given}$

A circle with centre O; PA and PB are two tangents to the circle drawn from an external point P.

To prove: PA = PB

$\bf \large \underline {Construction:}$

Join OA, OB, and OP.

◕we know that ◕

A tangent at any point of a circle is perpendicular to the radius through the point of contact.

$\bf OA⊥PA \\ </p><p></p><p> \bf \implies OB⊥PB \\ </p><p></p><p> \bf \implies In △OPA and △OPB \\ </p><p></p><p> \bf \implies \: ∠OPA=∠OPB (Using (1)) \\ </p><p></p><p> \bf \implies \: OA=OB (Radii of the same circle) \\ </p><p></p><p> \bf \implies OP=OP (Common side) \\ </p><p></p><p> \bf \implies \: Therefor △OPA≅△OPB (RHS congruency criterion) \\ </p><p></p><p> \bf \implies PA=PB$

(Corresponding parts of congruent triangles are equal)

☞Thus, it is proved that the lengths of the two tangents drawn from an external point to a circle are equal.

☞The length of tangents drawn from any external point are equal

$\bf \pink So \: statement \: is \: correct \:$