Question

In the circle centred at O, the tangents at A and B intersect at P. Prove the following:

(i)

the point P is equidistant from A and B

(ii)

the line OP bisects the line AB and the angle APB

(iii)

if the line OP cuts the line AB at Q, then OQ × OP = r², where r is the radius of the circle

Answer 1

Given: A circle with centre O

Tangents at points A and B intersect at point P.

(i)

Construction: Join AO, OB and OP.

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

AP is a tangent to the circle and OA is the radius of the circle.

∴∠OAP = 90°

Applying Pythagoras theorem in ΔOAP:

OP² = OA² + AP²

_⇒OA² = OP² − AP² … (1)

Similarly, we have OB² = OP² − BP² …(2)

OA = OB (Radii of the same circle)

From equations (1) and (2), we have:

OP² − AP² = OP² − BP²

⇒ AP² = BP²

⇒ AP = BP

Thus, point P is equidistant from A and B.

(ii)

Construction: Join AB.

In ΔOAP and ΔOBP:

PA = PB (Proved in part (1))

PO = PO (Common side)

OA = OB (Radii of the same circle)

As all sides of one triangle are equal to the corresponding sides of the other triangle, ΔOAP ≅ ΔOBP.

Corresponding parts of congruent triangles are congruent.

⇒∠AOP = ∠BOP …(1)

∠APO = ∠BPO

Thus, line segment OP bisects ∠APB.

In ΔAQO and ΔBQO:

OA = OB (Radii of the same circle)

∠AOP = ∠BOP (From equation (1))

OQ = OQ (Common side)

As two sides and included angle of one triangle are equal to the two sides and included angle of the other triangle, ΔAQO ≅ ΔBQO.

Corresponding parts of congruent triangles are congruent.

⇒ AQ = BQ

Thus, line segment OP bisects AB.

(iii)

We have shown in part (2) that ΔAQO ≅ ΔBQO.

∴ ∠AQO = ∠BQO (Corresponding parts of congruent triangles are congruent)

We know that sum of angles forming a linear pair are equal.

∴ ∠AQO + ∠BQO = 180°

⇒ ∠AQO = ∠BQO = 90°

In ΔAQO and ΔPBO:

∠AQO = ∠PBO = 90°

∠AOP = ∠BOP (Proved in part (1))

∴ ΔAQO ∼ ΔPBO (By angle angle criterion)