Angles in a Cyclic Quadrilateral

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

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

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.

Given: a cyclic quadrilateral ABCD

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°