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Topic 2/3
15 Flashcards in this deck.
Ptolemy’s Theorem:
In a cyclic quadrilateral $ABCD$, the product of the diagonals is equal to the sum of the products of opposite sides. Mathematically, this is expressed as: $$AC \cdot BD = AB \cdot CD + AD \cdot BC$$
Where:
Brahmagupta’s Formula:
The area $K$ of a cyclic quadrilateral with sides $a$, $b$, $c$, and $d$ is given by: $$K = \sqrt{(s-a)(s-b)(s-c)(s-d)}$$
Where $s$ is the semi-perimeter of the quadrilateral: $$s = \frac{a + b + c + d}{2}$$
Example: Given a cyclic quadrilateral $ABCD$ with sides $AB = 5$, $BC = 6$, $CD = 5$, and $DA = 6$, and one of the diagonals $AC = 7$, find the length of the other diagonal $BD$.
Solution:
Using Ptolemy’s Theorem: $$AC \cdot BD = AB \cdot CD + AD \cdot BC$$ Substituting the known values: $$7 \cdot BD = 5 \cdot 5 + 6 \cdot 6$$ $$7 \cdot BD = 25 + 36$$ $$7 \cdot BD = 61$$ $$BD = \frac{61}{7} \approx 8.71$$
Proof:
Consider a cyclic quadrilateral $ABCD$ with vertices lying on a circle. Extend sides $AB$ and $CD$ to meet at point $E$. Since $ABCD$ is cyclic, angles $\angle A$ and $\angle C$ subtend the same arc. Therefore, $\angle A + \angle C = 180^\circ$ because they are supplementary angles.
Problem: In cyclic quadrilateral $ABCD$, $AB = 8$, $BC = 15$, $CD = 7$, and $DA = 10$. Find the lengths of the diagonals $AC$ and $BD$.
Solution:
Applying Ptolemy’s Theorem: $$AC \cdot BD = AB \cdot CD + AD \cdot BC$$ $$AC \cdot BD = 8 \cdot 7 + 10 \cdot 15$$ $$AC \cdot BD = 56 + 150$$ $$AC \cdot BD = 206$$
Additionally, using Brahmagupta’s Formula to find the area $K$:
$$s = \frac{8 + 15 + 7 + 10}{2} = 20$$ $$K = \sqrt{(20-8)(20-15)(20-7)(20-10)} = \sqrt{12 \cdot 5 \cdot 13 \cdot 10} = \sqrt{7800} \approx 88.36$$Using the area formula related to diagonals: $$K = \frac{AC \cdot BD \cdot \sin \theta}{2}$$ Assuming $\theta$ is the angle between the diagonals, and since $ABCD$ is cyclic, $\theta = 90^\circ$, so $\sin \theta = 1$: $$88.36 = \frac{AC \cdot BD}{2}$$ $$AC \cdot BD = 176.72$$
However, from Ptolemy’s Theorem, $AC \cdot BD = 206$, which suggests a discrepancy. This indicates that the assumption $\theta = 90^\circ$ may not hold, and further trigonometric analysis is required to find the exact lengths of $AC$ and $BD$. This problem illustrates the need for integrating multiple concepts to solve advanced problems.
Derivation of Ptolemy’s Theorem:
Consider a cyclic quadrilateral $ABCD$ inscribed in a circle. By extending sides $AD$ and $BC$, intersecting at point $E$, we can apply similar triangles and proportions to derive: $$AC \cdot BD = AB \cdot CD + AD \cdot BC$$ This derivation relies on properties of similar triangles and the power of a point theorem.
Problem: In cyclic quadrilateral $ABCD$, $AB = 10$, $BC = 6$, $CD = 8$, and $DA = x$. If one of the diagonals $AC = 12$, find the value of $x$.
Solution:
Applying Ptolemy’s Theorem: $$AC \cdot BD = AB \cdot CD + AD \cdot BC$$ $$12 \cdot BD = 10 \cdot 8 + x \cdot 6$$ $$12 \cdot BD = 80 + 6x$$
To find another equation, use Brahmagupta’s Formula to relate the sides and the radius. However, additional information is required to solve for $x$. This problem illustrates the complexity of advanced cyclic quadrilateral problems, often requiring multiple theorems and formulas for a solution.
Aspect | Cyclic Quadrilateral | General Quadrilateral |
Definition | All four vertices lie on a single circle. | Vertices do not necessarily lie on a single circle. |
Sum of Opposite Angles | Each pair sums to $180^\circ$. | No specific sum. |
Ptolemy’s Theorem | Applies, providing a relationship between sides and diagonals. | Does not necessarily hold. |
Area Calculation | Can use Brahmagupta’s Formula. | Requires different methods, such as Bretschneider’s Formula. |
Symmetry | Often exhibits more symmetry. | Symmetry varies. |
To excel in topics involving cyclic quadrilaterals, always start by verifying if the quadrilateral is cyclic using the sum of opposite angles. Use mnemonic devices like "Ptolemy Protects Products" to remember that Ptolemy’s Theorem relates the products of the diagonals and sides. Practice drawing accurate circumcircles to better visualize properties and relationships. Additionally, regularly solve varied problems to become comfortable with applying Brahmagupta’s Formula and integrating multiple theorems. These strategies will enhance your problem-solving skills and boost your confidence in exams.
Cyclic quadrilaterals have fascinated mathematicians for centuries. Did you know that Ptolemy, an ancient Greek mathematician, first formulated Ptolemy’s Theorem over 2,000 years ago? Additionally, Brahmagupta’s Formula, which calculates the area of a cyclic quadrilateral, was discovered by the Indian mathematician Brahmagupta in the 7th century. These principles are not only theoretical but also apply to modern technologies such as computer-aided design and architecture, where precise geometric calculations are crucial.
Students often make errors when working with cyclic quadrilaterals. One common mistake is assuming that all quadrilaterals with opposite angles summing to $180^\circ$ are cyclic without verifying if the vertices lie on a single circle. Another frequent error is misapplying Ptolemy’s Theorem by forgetting that it only applies to cyclic quadrilaterals. For example, attempting to use the theorem on a general quadrilateral can lead to incorrect results. Additionally, neglecting to use Brahmagupta’s Formula correctly by not calculating the semi-perimeter accurately may result in wrong area calculations.