Question

pove that
The length of two tangent drawn
from an external point
to
a circle are equal
​

Answer 1

Answer:

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.

Answer 2

Answer:

☆ ANSWER ☆

LETS TAKE THE TANGENTS AP AND BP WHERE P IS THE EXTERNAL POINT OF THE CIRCLE WITH CENTER O.

JOIN OP.

LETS TAKE TRIANGLE AOP AND BOP.

< OAP = < OBP = 90 (TANGENT TO THE CIRCLE)

OP = OP ( COMMON )

OA = OB ( RADIUS )

THEREFORE

TRIANGLE AOP AND BOP ARE CONGRUENT BY RHS CONGRUENCE RULE.

THEREFORE,

C.P.C.T.

AP = BP

HENCE PROVED

HOPE IT HELPS