Lesson 3, Topic 4
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# Alternate Segment

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Theorem: The angle between the tangent to a circle and the chord through the point of contact is equal to the angle in the alternate segment.

### Proof – Method 1:

Note: ∠a is the angle between the tangent & chord of the major segment, similarly ∠x is the angle between the tangent & chord of the minor segment as shown below.

To prove: a° (the angle between the tangent and the chord) = b° (the angle in the alternate segment)

i.e, a° = b°

First, let’s draw radii from the centre of the circle to the points on the triangle and tangent as shown below.

What we now have is an isosceles triangle (because the radii are equal), which we can divide in half by drawing a line through the middle, as shown below.

What do we have?

x = x (because it’s an isosceles triangle)

y = y (because we divided the triangle in half)

a + x = 90° ……(i) (tangent perpendicular to radius)

Consider one half of the isosceles triangle:

90° + x + y = 180° (sum of angles in a triangle)

∴ x + y = 180° – 90°

x + y = 90° …..(ii)

Equating i & ii

a = 90° – x
y = 90° – x
∴ a = y

But we know that 2y = 2b (angle at the centre is twice the angle at the circumference)

Note: the angle at the centre is y + y = 2y and b is twice the angle at the centre.

2y = 2b

∴ b = y

If a = y and b = y

then a = b

### Proof – Method 2:

Given: a circle with ABC as a tangent at B and C. |BD| is a chord dividing the circle into two segments BPD and BRD.

BPD is the alternate segment to angle CBD. Similarly, BRD is the alternate segment to angle ABD.

Therefore, ∠BPD is the angle in the alternate segment to ∠CBD.

Also, ∠BRD is the angle in the alternate segment to ∠ABD.

To prove:

i. ∠BPD = ∠CBD

ii. ∠BRD = ∠ABD

### i. ∠BPD = ∠CBD

Construction:

Draw the diameter BQ join DQ

Proof:

i. ∠CBQ = 90° (tangent perpendicular to radius)

Then ∠CBD = 90° – ∠DBQ

Also BDQ = 90° (angle in a semicircle)

∠BQD = 90° – ∠DBQ (sum of angles in a triangle)

But ∠BQD = ∠BPD (angles in the same segment)

∠BPD = 90° – ∠DBQ

∴ ∠BPD = ∠CBD

### ii.∠BRD = ∠ABD

∠ABD = 180° − ∠CBD (angles on a straight line)

∠ABD = 180° − ∠BPD ——- i

∠BPD + ∠BRD = 180° (sum of opposite angles on a cyclic quadrilateral)

∠BRD = 180° − ∠BPD ——— ii

Equating i & ii

∠ABD = ∠BRD

### Example 1

1. TP and TQ are tangents to a circle with centre O and R is a point on the circumference of the circle as shown in the diagram below. If the angle PTQ = 45°, what is the value of ∠PRQ?

Solution:

Since |TQ| and |TP| are tangents, P then |TQ|=|TP|

∴ ∆PQT is isosceles

Thus ∠PQT = ∠TPQ = $$\frac{180\: – \: 45}{2}$$

∴ ∠TPQ = 67.5°

But ∠TPQ = ∠PRQ (angles in alternate segment)

∴ ∠PRQ = 67.5°

### Example 2

In the diagram, PTR is a tangent to the centre O. If $$\scriptsize \hat{TON} = 108^o$$, Calculate the size of angle PTN.

Solution:

108° + ∠OTN + ∠ONT = 180° (sum of angles in a triangle)

∴ ∠OTN = ∠ONT (base angles of isosceles triangle)

let ∠OTN = ∠ONT = x°

108° + x° + x° = 180°

∴ 2x° = 180° – 108°

2x° = 72°

x = $$\frac{72}{2}$$

x = 36°

∠OTN = 36°

∠OTP = 90° (angle between tangent and radius of circle)

∴ ∠PTN = ∠OTP + ∠OTN

∠PTN = 90° + 36°

∠PTN = 126°

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