# 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°.

**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°.