Ptolemys Theorem for Cyclic Quadrilaterals

Ptolemy Theorem is an important component of Euclidean Geometry. It is named after the Greek Mathematician and astronomer Ptolemy. The theorem is derived by him while applying to astronomy to supplement his table of chords and trigonometry for astronomy. In this article, we will discuss the Ptolemy’s theorem proof and Ptolemy’s theorem applications. Some Ptolemy's theorem examples will also be discussed for a better understanding of the topic. The theorem is mainly focused on the concept of cyclic quadrilaterals, so knowledge of cyclic quadrilaterals is essential for a better understanding of the theorem.

Table of Contents

• Ptolemy Theorem: Brief introduction

• History of Claudius Ptolemy

• Statement of Ptolemy Theorem

• Proof of Ptolemy Theorem

• Limitations of the Ptolemy Theorem

• Applications of the Ptolemy Theorem

• Solved Examples

• Important Formulas to Remember

• Important Points to Remember

History of Claudius Ptolemy

Claudius Ptolemy

Image Credit: Wikimedia

Name: Claudius Ptolemy

Born:  100 AD

Died: 170 AD

Field: Mathematics and Astronomy

Nationality: Roman

Statement of Ptolemy Theorem

Ptolemy Theorem states that when a quadrilateral is inscribed in a circle, then the product of diagonals of the quadrilateral is equal to the sum of the product of pairs of opposite sides.

Proof of Ptolemy Theorem

Ptolemy's Theorem

Let us consider $A B C D$ a cyclic quadrilateral.

Now we can see that the chord $B C$, the inscribed angles $\angle B A C=\ angle B D C$, and on $A B, \angle A D B=\angle A C B$.

Now, Construct $K$ on $A C$ such that $\angle A B K=\angle C B D$.

(Note that: $\angle A B K+\angle C B K=\angle A B C=\angle C B D+\angle A B D \Rightarrow \angle C B K=\angle A B D .$ )

Now, by common angles property,

$\triangle K B C \sim \triangle A B D$

Similarly,

$\triangle A B K$ is similar to $\triangle D B C$.

Thus, $\dfrac{|A K|}{|A B|}=\dfrac{|D C|}{|D B|} and \dfrac{|K C|}{|B C|}=\dfrac{|A D|}{|B D|}$ due to the similarities noted above:

{$\triangle A B K \sim \triangle D B C and \triangle K B C \sim \triangle A B D$}

So $|A K| \cdot|D B|=|A B| \cdot|D C|$, and $|K C| \cdot|B D|=|B C| \cdot|A D|$

Adding,

$\Rightarrow|A K| \cdot|D B|+|K C| \cdot|B D|=|A B| \cdot|D C|+|B C| \cdot|A D|$

Equivalently, $(|A K|+|K C|) \cdot|B D|=|A B| \cdot|C D|+|B C| \cdot|A D|$

But

$\Rightarrow|A K|+|K C|=|A C|$

So,

$\Rightarrow|A C| \cdot|B D|=|A B| \cdot|C D|+|B C| \cdot|D A|$

Hence Proved.

Limitations of the Ptolemy Theorem

• It is only applicable in the case of cyclic quadrilaterals.

• It only tells us about the sides and diagonal length relationship and doesn’t give any idea about the angles of the quadrilateral.

Applications of the Ptolemy Theorem

• Ptolemy's Theorem has a wide range of applications in astronomy. The discovery of the theorem is done for astronomy itself.

• It is used to find the value of sides or diagonals of cyclic quadrilateral using the formula stated in the theorem.

• It can be used to derive the Pythagoras theorem.

Solved Examples

1. In the given figure, $A B=10 \mathrm{~cm}, D C=5 \mathrm{~cm}, B C=20 \mathrm{~cm}, A D=15 \mathrm{~cm}$ and $A C=25 \mathrm{~cm}$, then find $D B$.

ABCD is a Cyclic Quadrilateral

Ans: By Ptolemy's Theorem, we have,

$D B \times A C=A B \times D C+A D \times B C$

Putting values, we get,

$\Rightarrow D B \times 25=10 \times 5+15 \times 20 $

$\Rightarrow 25 D B=50+300$

$25 D B=350$

$D B=\dfrac{70}{5}$

$D B=14 \mathrm{~cm}$

2. In the given figure, $A B=30 \mathrm{~cm}, D C=5 \mathrm{~cm}, B C=20 \mathrm{~cm}, A D=25 \mathrm{~cm}$ and $A C=25 \mathrm{~cm}$, then find $D B$.

Cyclic Quad ABCD

Ans: By Ptolemy's Theorem, we have,

$D B \times A C=A B \times D C+A D \times B C$

Putting values, we get,

$\Rightarrow D B \times 25=30 \times 5+25 \times 20 $

$\Rightarrow 25 D B=150+500 $

$\Rightarrow 25 D B=650 $

$\Rightarrow D B=\dfrac{130}{5} $

$\Rightarrow D B=26 \mathrm{~cm}$

3. In given figure, $A B=10 \mathrm{~cm}, D C=15 \mathrm{~cm}, B C=35 \mathrm{~cm}, A D=40 \mathrm{~cm}$ and $D B=25 \mathrm{~cm}$, then find $A C$.

To Find the Diagonal of Cyclic Quadrilateral

Ans: By Ptolemy's Theorem, we have,

$D B \times A C=A B \times D C+A D \times B C$

Putting values, we get,

$\Rightarrow A C \times 25=10 \times 15+40 \times 35 $

$\Rightarrow 25 A C=150+1400 $

$\Rightarrow 25 A C=1550 $

$\Rightarrow A C=\dfrac{310}{5} $

$\Rightarrow A C=62 \mathrm{~cm}$

Conclusion

We have discussed the detailed proof of Ptolemy Theorem and its applications in this article. Ptolemy Theorem forms a fundamental tool for cyclic quadrilaterals. In all, we can say that Ptolemy's Theorem is a fantastic theorem which helps us in solving problems of cyclic quadrilaterals easily.

Important Formulas to Remember

• For a cyclic quadrilateral ABCD, with AC and BD be the diagonals we have, $\left| {AC} \right|.\left| {BD} \right| = \left| {AB} \right|.\left| {CD} \right| + \left| {BC} \right|.\left| {DA} \right|$.

Important Points to Remember

• A quadrilateral is said to be cyclic if all the vertices of the quadrilateral lie on the circumference of the circle.

List of Related Articles

• Cyclic Quadrilaterals

• Circles