Question

Two tangents PA and PB are drawn to a circle with centre O from an external point P. Prove that Angle APB =2 Angle OAB​

Answer 1

PA & PB are the tangents to a circle, with Centre O from a point P outside it.

To Prove:

∠APB = 2∠AOB

Proof:

Let ∠APB= x°

We know that the tangents to a circle from an external point are equal in length so PA= PB.

PA =PB

∠PBA = ∠PAB

[Angles opposite to the equal sides of a triangle are equal.]

∠APB+ ∠PBA +∠PAB= 180°

[Sum of the angles of a triangle is 180°]

x° + ∠PAB +∠PAB = 180°

[∠PBA = ∠PAB]

x° + 2∠PAB = 180°

∠PAB =½(180° - x°)

∠PAB =90° - x°/2

∠OAB +∠PAB=90°

∠OAB =90° - ∠PAB

∠OAB =90° - (90° - x°/2)

∠OAB =90° - 90° + x°/2

∠OAB = x°/2

∠OAB = ∠APB /2

∠OAB = 1/2∠APB

∠APB = 2∠OAB