Question

form external point p two tangents PA and PB are dramn to the circle with center O prove that OP is the perpendicular bisector of AB​

Answer 1

Answer:

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

∠ OAP = ∠ OBP = 90°

Now, ∠ OAP + ∠ APB + ∠ OBP + ∠ AOB = 360° [Angle sum property of quadrilaterals]

⟹ 90° + ∠ APB + 90° + ∠ AOB = 360°

⟹ ∠ AOB = 360° - 180° - ∠ APB = 180° - ∠ APB.... (1)

Now, in Δ OAB, OA is equal to OB as both are radii.

⟹ ∠ OAB = ∠ OBA [In a triangle, angles opposite to equal sides are equal]

Now, on applying angle sum property of triangles in Δ AOB,

We obtain ∠ OAB + ∠ OBA + ∠ AOB = 180°

⟹ 2∠ OAB + ∠ AOB = 180°

⟹ 2 ∠ OAB + (180° - ∠ APB) = 180° [Using (1)]

⟹ 2 ∠ OAB = ∠ APB

Thus, the given result is proved

Answer 2

Answer:

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