Cyclic quadrilateral is a special kind of quadrilateral inscribed in a circle. Inscribe means that the vertices of quadrilateral are on the circumference of the circle. A square, rectangle, isosceles trapezoid are some of cyclic quadrilaterals. Below are the some facts about this special figure.

1. The opposite angle of cyclic quadrilateral is supplementary.

$\angle{A}+\angle{C}=180$°

$\angle{B}+\angle{D}=180$°

2. Ptolemy’s Theorem. This theorem is a special case of Ptolemy’s Inequality. He formulated the following equation relating the sides and diagonals.

$d_1d_2=ac+bd$

3. Diagonals as Cords. Using the basic theorem in geometry, we can relate the diagonals with respect to their intersections as follows.

$AO(OC)=OD(OB)$

4. Brahmagupta’s Formula. Brahmagupta of India derived a formula to find the area of cyclic quadrilateral.

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

Where S is the semiperimeter

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

Use the facts above to solve for the following problems.

Problem 1:  In a cyclic quadrilateral $ABCD$, If $\angle{A}=105$°. Find the measure of $\angle{C}$

Solution:

Using the first fact, opposite angles are supplementary.

$\angle{A}+\angle{C}=180$°

$105$°$+\angle{C}=180$°

$\angle{C}=(180-105)$°

$\angle{C}=75$°

Problem 2: In cyclic quadrilateral $ABCD$, $\angle{C}=\angle{A}=90$°. If $\overline{BD}=5$, and two opposite sides have length 2 and 3. What is the area of quadrilateral?

Solution:

Since $\Delta BCD$ and $\Delta ABD$ are right triangle and $\overline{BD}$ is the hypotenuse of both triangles, we can find sides $\overline{AD}$  and   $\overline{BC}$  by Pythagorean Theorem.

$\overline{BD}^2=\overline{BC}^2+\overline{CD}^2$

$5^2=\overline{BC}^2+3^2$

$\overline{BC}^2=25-9=16$

$\overline{BC}=4$

$\overline{BD}^2=\overline{AB}^2+\overline{AD}^2$

$5^2=2^2+\overline{AD}^2$

$\overline{AD}^2=25-4=21$

$\overline{AD}=\sqrt{21}$

The area can be found using Brahmagupta’s formula or using the fact that both triangles are right triangle.

$Area_{quadrilateral}=Area_{\Delta BCD}+Area_{\Delta ABD}$

$Area_{quadrilateral}=\displaystyle\frac{\overline{BC}\cdot \overline{CD}}{2}+\displaystyle\frac{\overline{AB}\cdot \overline{AD}}{2}$

$Area_{quadrilateral}=\displaystyle\frac{1}{2}(3\cdot 4+2\cdot \sqrt{21})$

$Area_{quadrilateral}=6+\sqrt{21}$ sq. units