Facts About Cyclic Quadrilateral

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.

Cyclic Quadrilateral

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:

Draft the quadrilateral as follows,

cyclic quadrilateralSince \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

 

 

 

Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.
Dan

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