Table of Contents

Last modified on August 3rd, 2023

A cyclic quadrilateral is a quadrilateral with all its four vertices or corners lying on the circle. It is thus also called an inscribed quadrilateral. When any four points on the circumference of a circle are joined, they form the vertices of a cyclic quadrilateral.

- Has four vertices that lie on the circumference of a circle. The four vertices make four angles; in a cyclic quadrilateral, ABCD, A, B, C, D are the vertices making angles ∠DAB, ∠ABC, ∠BCD, and ∠CDA
- Opposite angles add up to 180°; so ∠DAB + ∠BCD = 180° and ∠ABC + ∠CDA = 180°
- The exterior angle formed when any one side is extended is equal to the opposite interior angle; ∠DCE = ∠DAB

In a cyclic quadrilateral, the sum of the opposite angles is 180°. In other words, the pair of opposite angles in a cyclic quadrilateral is supplementary. The formula is given below:

**In the cyclic quadrilateral ABCD, find ∠DAB and ∠ABC when ∠BCD = 110° and ∠CDA= 70°.**

Solution:

As we know,

∠DAB + ∠BCD = 180° and ∠ABC + ∠CDA = 180°, here ∠BCD = 110° and ∠CDA= 80°

Now, ∠DAB + 110° = 180°

∠DAB = 180° – 110°

= 70°

Similarly,

∠ABC + 80° = 180°

∠ABC = 180° – 80°

= 100°

They are the line segments joining opposite vertices or corners of the cyclic quadrilateral. There are two diagonal (d_{1} and d_{2}) in a cyclic quadrilateral. The formula is given below:

**Find the diagonals of a cyclic quadrilateral whose sides are 3 m, 4 m, 6m, and 8 m.**

Solution:

As we know,

Diagonal (d_{1}) = √(ac + bd) (ad + bc)/ab +cd

= √[(3 x 6) + (4 x 8)] [(3 x 8) + (4 x6)]/[(3 x 4) + (6 x 8)] m

= √[(18 + 32) (24 + 24)]/[12 + 48] m

= √2400/60 m

= √40 m

= 6. 3245 m

Similarly,

Diagonal (d_{2}) = √(ac + bd) (ab + cd)/ad + bc

= √[(3 x 6) + (4 x 8)] [(3 x 4) + (6 x 8)]/ [(3 x 8) + (4 x 6)] m

= √[(18 + 32) (12 + 48)]/[24 + 24] m

= √3000/48 m

= √62.5 m

= 7.9056 m

Also known as circumradius, it is a straight line drawn from the center to the circumference of the circle in a cyclic quadrilateral. Parameshvara’s formula for finding circumradius is given below:

**Find the circumradius of the cyclic quadrilateral ABCD with center O whose sides measure 2 m, 4 m, 6m, and 10 m.**

Solution:

As we know,

Circumradius (R) = 1/4√(ab + cd) (ac + bd) (ad + bc)/ (s-a) (s-b) (s-c) (s-d), here a = 2 cm, b = 4 cm, c = 6 cm, and d = 10 cm

In this cyclic quadrilateral, s = ½ (2 cm + 4 cm + 6 cm + 10 cm) = 11 cm

Since, R = 1/4√(ab + cd) (ac + bd) (ad + bc)/ (s-a) (s-b) (s-c) (s-d)

= 1/4√[{(2 X 4) + (6 X 10) } {(2 X 6) + (4 x 10)} {(2 x 10) + (4 x 6)}]/[(11 – 2)(11 – 4)(11 – 6)(11 – 10)] cm

= 1/4√[{8 + 60} {12 + 40} {20 + 24}]/[9 x 7 x 5 x 1] cm

= 1/4√1,55,584/315 cm

= 1/4√493.9174 cm

= 5.556 cm

It is the total space enclosed by the cyclic quadrilateral. Brahmagupta’s formula for finding the area is given below:

**Find the area of a cyclic quadrilateral whose sides are 6 cm, 8 cm, 10 cm, and 12 cm.**

Solution:

As we know,

Area (A) = √(s-a) (s-b) (s-c) (s-d), here a = 2 cm, b = 4 cm, c = 6 cm, and d = 8 cm

In this cyclic quadrilateral, s = ½ (2 cm + 4 cm + 6 cm + 8 cm) = 10 cm

Since, A = √(s-a) (s-b) (s-c) (s-d)

= √(10-2)(10-4)(10-6)(10-8) cm2

=19. 5959 cm2

There are two important theorems that prove a cyclic quadrilateral. They are given below with their mathematical proofs.

Prove that opposite angles of a cyclic quadrilateral are supplementary

To prove:

∠ABC + ∠CDA = 180° and ∠DAB + ∠BCD = 180°

Proof:

Let ABCD be a cyclic quadrilateral and ‘O’ be the center of the circle.

Join OB and OD.

a) ∠DAB = ½ ∠BOD (By theorem: The angle extended by an arc at the center is double the angle on the circle)

b) Similarly, ∠BCD = ½ reflex ∠BOD

c) Again, ∠DAB + ∠BCD = ½ reflex ∠BOD + ½ reflex ∠BOD

Adding (a) and (b) we get,

∠DAB + ∠BCD = ½(∠BOD + reflex ∠BOD)

∠DAB + ∠BCD = ½ x 360°

∠DAB + ∠BCD = 180°

Similarly,

∠ABC + ∠CDA = 180°

Hence Proved

Prove Ptolemy’s Theorem

To prove:

AC x BD = AD x CB + DC x AB

Proof:

Let ABCD be a cyclic quadrilateral, and ‘P’ be a point on the cord AC such that ∠ADP = ∠CDB

a) ∠ADP + ∠CDP = ∠ADC

b) ∠CDB + ∠ADB = ∠ADC

Adding (a) and (b) we get,

∠CDP = ∠ADB

c) Again, ∆ADP ∼ ∆DBC and ∆PDC ∼ ∆ADB (By theorem: Similarity of triangles)

Thus we get,

i) |AP|/|AD| = |BC|/|BD|

ii) |PC|/|DC| = |AB|/|DB|

Cross-multiplying within (i) and (ii) we get,

iii) |AP| x |BD| = |AD| x |BC|

iv) |PC| x |DB| = |DC| x |AB|

Now, adding (iii) and (iv) we get,

|AP| x |BD| +|PC| x |DB| = |AD| x |BC| + |DC| x |AB|

Similarly, (|AP| + |PC|) |BD| = |AD| x |BC| + |DC| x |AB|

But, |AP| + |PC| = |AC|

Thus, |AC| x |BD| = |AD| x |CB| + |DC| x |AB|

Hence Proved

Last modified on August 3rd, 2023