Prove that “The lengths of the two tangent segments to a circle drawn from an external point are equal.”

#### Solution 1

Given: A circle with centre O and an external point P are given.

AP and BP are the two tangents drawn from an external point P.

To prove: AP = BP

Construction: Draw seg OA, seg OB and seg OP.

Proof: In ΔOBP and ΔOAP,

OA = OB … (Radii of the same circle)

OP = OP … (Side common to both the triangles)

∠OAP = ∠OBP = 90° … (tangent is perpendicular to the radius at the point of contact)

ΔOBP ≅ ΔOAP … (By R.H.S)

∴ AP = BP … (corresponding sides of congruent triangles)

Thus, the lengths of two tangent segments to a circle drawn from an external point are equal.

#### Solution 2

Given: O is the centre of the circle and P is a point in the exterior of the circle. A and B are the points of contact of the two tangents from P to the circle.

To Prove: PA = PB

Construction: Draw seg OA, seg OB and seg OP.

Proof: Line AP ⊥ radius OA and line BP ⊥ radius OB ... (Tangent perpendicular to radius)

∴ `anglePAO = anglePBO = 90^@`

In right-angled triangles `triangle OAP` and `triangleOBP`

hypotenuse OP ≅ hypotenuse OP ...(Common side)

seg OA ≅ seg OB ...(Radii of the same circle)

`:.triangleOAP ≅ triangleOBP` ...(Hypotenuse-side of theorem)

∴ seg PA ≅ seg PB ...(c.s.c.t.)

∴ PA = PB