In Figure 10.13, XY and X′ Y′ are two parallel tangents to a circle with centre O and another tangent AB with the point of contact C intersecting XY at A and X′ Y′ at B. Prove that ∠AOB = 90°

Solution:

Draw a line between points O and C.

In ΔOPA and ΔOCA

OP = OC (Radii of the circle)

AP = AC (The lengths of tangents drawn from an external point to a circle are always equal.)

AO = AO (Common)

By SSS congruency, ΔOPA ≅ ΔOCA

SSS congruence rule: If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

Therefore, ∠POA = ∠AOC ---------- (1)

Similarly, ΔOCB ≅ ΔOQB

Therefore, ∠COB = ∠BOQ ----------- (2)

PQ is a diameter, hence a straight line and ∠POQ = 180°

But ∠POQ = ∠POA + ∠AOC + ∠COB + ∠BOQ

∴ ∠POA + ∠AOC + ∠COB + ∠BOQ = 180°

2∠AOC + 2∠COB = 180° [From equation (1) and (2)]

∴ ∠AOC + ∠COB = 90°

From the figure,

∠AOC + ∠COB = ∠AOB

∴ ∠AOB = 90°

Hence Proved ∠AOB = 90°.

☛ Check: NCERT Solutions for Class 10 Maths Chapter 10

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Video Solution:

In Figure 10.13, XY and X′ Y′ are two parallel tangents to a circle with centre O and another tangent AB with the point of contact C intersecting XY at A and X′ Y′ at B. Prove that ∠AOB = 90°
 

NCERT Solutions Class 10 Maths Chapter 10 Exercise 10.2 Question 9

Summary:

If XY and X′ Y′ are two parallel tangents to a circle with centre O and another tangent AB with the point of contact C intersecting XY at A and X′ Y′ at B, we proved that ∠AOB = 90°.

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☛ Related Questions:

• Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
• Prove that the parallelogram circumscribing a circle is a rhombus.
• A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.
• Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.