

# Proofs of triangle properties: Pythagorean theorem, angle sum

To begin with, let's consider the Pythagorean theorem. It states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. Mathematically, it can be written as:

$$a^2 + b^2 = c^2$$

where $a$ and $b$ are the lengths of the two shorter sides, and $c$ is the length of the hypotenuse.

Now, let's prove this theorem using the method of exhaustion. We'll start by drawing a square with side length $c$. Inside this square, we'll draw a right-angled triangle with sides $a$ and $b$. The area of this triangle will be $\frac{1}{2}ab$.

Next, we'll draw a square with side length $b$. Inside this square, we'll draw a right-angled triangle with sides $a$ and $c$. The area of this triangle will be $\frac{1}{2}ac$.

Now, let's compare the areas of these two triangles. We know that the area of the first triangle is $\frac{1}{2}ab$, and the area of the second triangle is $\frac{1}{2}ac$. These areas are equal, so we can set them equal to each other:

$$\frac{1}{2}ab = \frac{1}{2}ac$$

Now, let's multiply both sides of the equation by $2$ to get rid of the fractions:

$$ab = ac$$

This equation tells us that the product of the lengths of the two shorter sides of the triangle is equal to the square of the length of the hypotenuse. Rearranging the equation, we get:

$$a^2 + b^2 = c^2$$

which is the Pythagorean theorem.

Here's an example of the Pythagorean theorem in action. Let's say we have a right-angled triangle with sides $3$, $4$, and $5$. Using the Pythagorean theorem, we can calculate the length of the hypotenuse:

$$5^2 = 3^2 + 4^2$$

$$25 = 9 + 16$$

$$25 = 25$$

As we can see, the equation is true, and the length of the hypotenuse is indeed $5$.

## Exercise
Prove the Pythagorean theorem using the method of exhaustion.

Next, let's consider the sum of the angles in a triangle. In any triangle, the sum of the angles is always $180^\circ$. This is known as the angle sum theorem.

To prove this theorem, let's consider a triangle with angles $A$, $B$, and $C$. We can draw a circle with the same circumference as the triangle. This circle will divide the triangle into three sectors, each with an angle equal to one of the triangle's angles.

Now, let's rotate the triangle so that the angle $A$ is at the top. When we do this, the angle $A$ will be opposite the largest sector, and the angles $B$ and $C$ will be opposite the two smaller sectors.

Since the sum of the angles in each sector is $180^\circ$, the sum of the angles in the triangle is:

$$A + B + C = 180^\circ$$

This proves the angle sum theorem.

Here's an example of the angle sum theorem in action. Let's say we have a triangle with angles $30^\circ$, $60^\circ$, and $90^\circ$. Using the angle sum theorem, we can calculate the sum of the angles:

$$30^\circ + 60^\circ + 90^\circ = 180^\circ$$

$$180^\circ = 180^\circ$$

As we can see, the equation is true, and the sum of the angles in the triangle is indeed $180^\circ$.

## Exercise
Prove the angle sum theorem using the method of drawing a circle.

# Real-world examples and applications of triangles: architecture, art, optics

Triangles are used in various fields of architecture, art, and optics. Let's consider some examples.

In architecture, triangles are used to create intricate patterns and designs. For example, in Islamic architecture, the use of triangles is common. The shape of the mihrab (a niche indicating the direction of Mecca) in a mosque is often a triangle.

In art, triangles are used to create balance and harmony in a composition. For example, in paintings and drawings, artists often use triangles to create a sense of depth and perspective.

In optics, triangles are used to calculate the focal length of lenses. The focal length is the distance between the lens and the image formed on a screen. It can be calculated using the following formula:

$$f = \frac{1}{2} \cdot \frac{d}{u}$$

where $f$ is the focal length, $d$ is the distance between the lens and the object, and $u$ is the distance between the lens and the image.

Here's an example of calculating the focal length of a lens. Let's say we have a lens with a distance of $10$ cm between the lens and the object, and a distance of $20$ cm between the lens and the image. Using the formula above, we can calculate the focal length:

$$f = \frac{1}{2} \cdot \frac{10}{20}$$

$$f = \frac{5}{10}$$

$$f = \frac{1}{2}$$

As we can see, the focal length of the lens is $0.5$ cm.

## Exercise
Calculate the focal length of a lens using the formula.

# Solving geometric problems using triangle properties

Triangles are often used to solve geometric problems. Let's consider some examples.

In a triangle with sides $3$, $4$, and $5$, we can use the Pythagorean theorem to find the length of the hypotenuse.

$$5^2 = 3^2 + 4^2$$

$$25 = 9 + 16$$

$$25 = 25$$

As we can see, the equation is true, and the length of the hypotenuse is indeed $5$.

In a triangle with angles $30^\circ$, $60^\circ$, and $90^\circ$, we can use the angle sum theorem to find the sum of the angles.

$$30^\circ + 60^\circ + 90^\circ = 180^\circ$$

$$180^\circ = 180^\circ$$

As we can see, the equation is true, and the sum of the angles in the triangle is indeed $180^\circ$.

## Exercise
Solve the following problem using triangle properties:

In a triangle with sides $7$, $8$, and $9$, find the length of the hypotenuse.

# Applications of triangles in physics: moment of inertia, light refraction, optics

Triangles are used in various fields of physics. Let's consider some examples.

In physics, triangles are used to calculate the moment of inertia. The moment of inertia is a measure of the resistance of a body to rotation. It can be calculated using the following formula:

$$I = \frac{1}{2} \cdot m \cdot r^2$$

where $I$ is the moment of inertia, $m$ is the mass of the object, and $r$ is the distance from the object's center of mass to a point on the object.

In optics, triangles are used to calculate the angle of light refraction. The angle of light refraction is the angle at which light is refracted when passing from one medium to another. It can be calculated using Snell's law:

$$\frac{\sin \theta_1}{\sin \theta_2} = \frac{n_1}{n_2}$$

where $\theta_1$ is the angle of incidence (the angle at which the light hits the surface), $\theta_2$ is the angle of refraction (the angle at which the light is refracted), $n_1$ is the refractive index of the first medium, and $n_2$ is the refractive index of the second medium.

Here's an example of calculating the moment of inertia of a disk. Let's say we have a disk with a radius of $2$ cm and a mass of $1$ kg. Using the formula above, we can calculate the moment of inertia:

$$I = \frac{1}{2} \cdot 1 \cdot (2)^2$$

$$I = \frac{1}{2} \cdot 4$$

$$I = 2$$

As we can see, the moment of inertia of the disk is $2$ kg-cm$^2$.

## Exercise
Calculate the moment of inertia of a disk using the formula.

# Advanced topics: triangle inequalities, barycentric coordinates, Euler's theorem

Triangles also have various advanced properties and concepts. Let's consider some examples.

In geometry, triangle inequalities are used to prove that certain properties hold for triangles. For example, the triangle inequality theorem states that in any triangle, the sum of the lengths of any two sides is always greater than the length of the third side.

In analytic geometry, barycentric coordinates are used to represent points in a triangle. Barycentric coordinates are a set of three numbers that can be used to represent any point inside a triangle. They are calculated using the following formula:

$$(u, v, w) = \frac{x}{a} \cdot (a, b, c) + \frac{y}{b} \cdot (b, c, a) + \frac{z}{c} \cdot (c, a, b)$$

where $(x, y, z)$ is the point in the triangle, and $(a, b, c)$ are the lengths of the sides.

In number theory, Euler's theorem is used to calculate the sum of the divisors of a number. The sum of the divisors of a number is the sum of all the numbers that divide that number. Euler's theorem can be used to calculate this sum using the following formula:

$$\sigma(n) = \sum_{d \mid n} d$$

where $\sigma(n)$ is the sum of the divisors of $n$.

## Exercise
Calculate the sum of the divisors of a number using Euler's theorem.