From a point Q, two tangents QA and QB are drawn to a circle with centre O. If OQ is diameter of the circle, Show AQB is an isosceles triangle.

Given: Tangents QA and QB are drawn to a circle with centre O.

To prove: AQB is isosceles triangle.

Theorem Used:

1) Tangent to a circle at a point is perpendicular to the radius through the point of contact.

2) The length of two tangents drawn from an external point are equal.

Explanation:

Join OA and OB.

We have OQ = Diameter

As OP = radius

⇒ OQ = OP + PQ

⇒ PQ = OQ – OP

⇒ PQ = radius

So, OP = PQ = radius

Since AQ is tangent to the radius OA,

From the theorem stated,

AQ ⊥ OA.

Thus, OQ is hypotenuse of ΔOAQ.

As mid-point on hypotenuse of a right-angled triangle is equidistant from the vertices.

∴ OA = AP = OP

⇒ OAP is an equilateral triangle.

⇒ ∠AOP = 60°

In ΔOAQ,

∠O + ∠A + ∠Q = 180° (angle sum property)

⇒ 60° + 90° + ∠Q = 180°

⇒ 150° + ∠Q = 180°

⇒ ∠Q = 30°

So, ∠AQO = 30°

∴ ∠AQB = 2∠AQO

⇒ ∠AQB = 60°

As Q is external point.

Also, QA and QB are tangents.

By theorem 2) stated above,

QA = QB

∠QBA =∠QAB (angles opp. to equal sides)

But ∠AQB = 60°

In ΔAQB,

∠AQB + ∠QBA +∠QAB = 180°

⇒ 60°+ 2∠QBA = 180°

⇒ 2∠QBA = 120°

⇒ ∠QBA = 60°

⇒ ∠QBA =∠QAB = 60°

As ∠AQB = ∠QBA =∠QAB =60°

ΔAQB is equilateral.

Hence proved