**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.

**Given: **A cyclic quadrilateral ABCD

**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**

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.

**Given: **a cyclic quadrilateral ABCD

**To prove:** ∠DAB = ∠ACE

**Construction:** Produce side BC to E

### Proof:

x_{1} + y = 180° **(opposite angles of a cyclic quadrilateral)**

x_{2} + y = 180° **(angles on a straight line)**

x_{1} = 180° – y

x_{2} = 180° – y

∴ x_{1} = x_{2}

### 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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