Question

32. Prove that the lengths of tangents drawn from an external point to a cirde are equal​

Answer 1

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..

Answer 2

Answer:

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

OA PA and OB PB ... (1)

In OPA and OPB:

OAP = OBP (Using (1))

OA = OB (Radii of the same circle)

OP = OP (Common side)

Therefore, OPA OPB (RHS congruency criterion)