Cyclic Quadrilateral

A cyclic quadrilateral is a four-sided polygon where all the corners touch the edge of a circle. The circle here is referred to as the circumcircle. Its centre is the circumcentre. There is a special place for cyclic quadrilaterals in mathematics because of their beautifully consistent arc and angle relationships.

These shapes can be found in engineering blueprints, astronomical calculations, and even navigation systems. Their properties help solve real-world problems. Mastering them opens the door to understanding the connection between polygons and circles and tackling tricky geometry puzzles. Let’s take a deeper look at them:

1.0What is a Cyclic Quadrilateral?

A cyclic quadrilateral is a quadrilateral with four vertices that is tucked inside a circle, with each corner resting inside the edge of the circle. The circle is called the circumcircle, and its centre is known as the circumcentre. Circumradius is the distance from the centre to the edge of the circle.

Cyclic in this refers to the Greek word “kuklos,” which means wheel or circle. The word quadrilatera comes from the Latin words “quadri” (four) and “latus” (side).

A few examples of cyclic quadrilaterals are squares, rectangles, antiparallelogram, or isosceles trapezoids.

2.0Properties of Cyclic Quadrilateral

Thanks to the properties of cyclic quadrilateral, it is easy to recognise them, making them invaluable in solving geometry problems:

1. All vertices are on the circle and are called concyclic points.
2. Four sides are the cords, with each side acting as the circumcircle’s chord.
3. The exterior angle is equal to the opposite interior angle, which is also the cyclic quadrilateral theorem.
4. Ptolemy’s relationship - The product of the diagonals is equal to the sum of the products of opposite sides: p x q = (a x c) + (b x d)
5. The four perpendicular bisectors intersect at the same point, the circumcentre.
6. The opposite angles in cyclic quadrilateral are supplementary.

3.0Area of a Cyclic Quadrilateral

To find the area of a cyclic quadrilateral, you have to use Brahmagupta’s formula:

$Area=\sqrt{(s-a)(s-b)(s-c)(s-d)}$

where, $s=\frac{a+b+c+d}{2}$

Fun fact: For students who can see the resemblance of Brahmagupta’s formula to Heron’s formula, the latter is just a special case of the former, when one side becomes zero.

4.0Theorems on Cyclic Quadrilateral

Other than the opposite-angles rule, there are two theorems related to cyclic quadrilaterals that you should know about:

Ptolemy’s Theorem: For a cyclic quadrilateral ABCD: If a = AB, b = BC, c = CD, d = DA, and diagonals p = AC, q = BD, then: p × q = (a × c) + (b × d)

Brahmagupta’s Theorem: The area K of a cyclic quadrilateral is: K = (s - a)(s - b)(s - c)(s - d), where s is the semi-perimeter.

5.0Example Problems of Cyclic Quadrilaterals

Example 1 – Finding an Angle

Problem: In PQSR, a cyclic quadrilateral, triangle PQR is equilateral. Find ∠QSR.

Solution:

In △PQR, ∠QPR=60∘

Opposite angles in a cyclic quadrilateral are supplementary:

∠QSR + ∠QPR = 180∘

∠QSR + 60∘ = 180∘

∠QSR = 120∘

Answer: ∠QSR=120.

Example 2 – Solving for a Variable

Problem: The angles of a cyclic quadrilateral are: (4y+2)∘, (y+20)∘, (5y−2)∘, and 7y∘. Find y.

Solution:

Sum of angles in a quadrilateral = 360∘:

(4y+2) + (y+20) + (5y−2) + 7y = 360

17y + 20 = 360 

17y = 340

y = 20

Answer: y = 20.

Example 3 – Using Ptolemy’s Theorem

Problem: In cyclic quadrilateral ABCD: AB = 5, BC = 7, CD = 4, DA = 6. Find AC × BD.

Solution:

According to the Ptolemy’s theorem:

AC × BD = (AB × CD) + (BC × DA)

AC x BD = (5 x 4) + (7 x 6)

AC x BD = 20 + 42

AC x BD = 62

Answer: Product of diagonals = 62.