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
