Lesson 3, Topic 1
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# Angles in a Cyclic Quadrilateral

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Definition: A cyclic quadrilateral is a four-sided polygon inscribed in a circle. All four vertices of the quadrilateral lie on the circumference of the circle.

Theorem 1: The opposite angles of a cyclic quadrilateral are supplementary.

To prove: ∠DAB + DCB = 180°

Construction: Join radii OD and OB to the centre O of the circle.

### Proof:

From the lettering in the given diagram:

reflex ∠DOB = 2x (angle at centre is twice the angle at circumference)

obtuse ∠DOB = 2y (angle at centre is twice the angle at circumference)

2x + 2y = 360° (angles at a point)

dIvide both sides by 2

$$\frac{2x\:+\:2y}{2} = \frac{360}{2} \\ \scriptsize \therefore \: x \: + \: y = 180^o$$

but x = ∠DAB
and y = DCB

∴ ∠DAB + DCB = 180°

Theorem 2: The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

To prove: ∠DAB = ∠ACE

Construction: Produce side BC to E

### Proof:

x1 + y = 180° (opposite angles of a cyclic quadrilateral)

x2 + y = 180° (angles on a straight line)

x1 = 180° – y

x2 = 180° – y

∴ x1 = x2

### Example 1

In the given problem, PQSR is a cyclic quadrilateral, and ΔPQR is an equilateral triangle, then find the measure of ∠QSR.

Solution:

Given, △PQR is an equilateral triangle. (angles in equilateral angle = 60° each)

∴ ∠QPR = 60°. Now, we know that opposite angles are supplementary.

∴ ∠QSR + ∠QPR = 180° (opposite angles of a cyclic quadrilateral)

From the above two equations we have,

∠QSR + 60° = 180°

∠QSR = 180° − 60°

∠QSR = 120°

∴ The measure of ∠QSR = 120°

### Example 2

Calculate the value of p and q, giving reasons for every step.

Solution:

p = ∠DBC = 65° (angles in the same segment)

∠DAB = 80° (exterior∠ of cyclic quadrilateral ABCD = interior opposite ∠)

q + ∠DAB + 45° = 180° (sum of ∠s in △DAB)

q + 80° + 45° = 180°

q = 180° – 125°

q = 55°

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