Question

0.22. Prow that the lengths of the tangents drawn from the external point to a circle are equal
​

Answer 1

Answer:

It is known that a tangent at any point of a circle is perpendicular to the radius through the point of contact. Therefore triangle OPA is congruent to triangle OPB by RHS criterion. Thus, it is proved that the lengths of the two tangents drawn from an external point to a circle are equal

Step-by-step explanation:

please make me Brainlist please please please please please please please

Answer 2

Step-by-step explanation:

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

Construction: Join OA, OB, and OP.

It is known that a tangent at any point of a circle is perpendicular to the radius through the point of contact.

OA⊥PA

OB⊥PB

In △OPA and △OPB

∠OPA=∠OPB (Using (1))

OA=OB (Radii of the same circle)

OP=OP (Common side)

Therefor △OPA≅△OPB (RHS congruency criterion)

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.

So statement is correct..