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Answer 1

Answer:

Solve : Construction : O, P; O, R; O, Q are joined.

Proof : Now, AQ and PQ are the two tangents of the circle with centre O at the points A and P case of tangent.

∴∠AOQ = ∠POQ

Similarly, in PR and BR , it can be proved that, ∠PQR = ∠ BOR

∴ ∠AOQ + ∠BOR =∠POQ + ∠POR

Again, (∠AOQ + ∠BOR) +(∠POQ +∠POR) = 180° [ $\because$ ∠AOB = 180°]

or, (∠POQ + ∠POQ) + (∠POQ +∠POR) = 180°

or, 2(∠POQ + ∠POR) = 180°

or, ∠POQ + ∠POR = 90°

or, ∠POQ + ∠POR = 90°

or, ∠QOR = 90°

∴ ∆QOR is a right angled triangle.

Hence, Proved

Answer 2

EXPLANATION.

AB is a diameter of circle with center o.

P is any point on circle.

p intersect the two tangent drawn through the point A, B at the point Q and R.

Radius of circle = r.

As we know that,

The length of a tangent to a circle drawn from the external point of the circle is equal and subtend equal angle at the center of the circle.

⇒ QA = QP.

⇒ RP = RB.

⇒ ∠AOQ = ∠POQ.

⇒ ∠POR = ∠BOR.

⇒ ∠AOQ = ∠POQ = 1/2∠AOP.

⇒ ∠POR = ∠BOR = 1/2∠BOP.

In ΔQOR,

⇒ ∠QOR = ∠POQ + ∠POR.

⇒ ∠QOR = 1/2∠AOB.

As we know that,

⇒ ∠AOB = 180°.

⇒ ∠QOR = 1/2(180°).

⇒ ∠QOR = 90°.

⇒ OP ⊥ QR.

⇒ ΔPOR ≅ ΔPOQ.

By Thales Theorem,

⇒ PQ/OP = OP/PR.

⇒ PQ.PR = OP².

⇒ PQ.PR = r².

HENCE PROVED.