#### Triangles

*Similar Triangles:*

If you have any triangle, and another one that looks identicle except that it is either larger or smaller (i.e. looks "blown-up" or "shrunk-down"), then those two triangles are *similar*. The mathematical definition for *similar triangles* is that they both have correponding angles that are equal, while the lengths of the corresponding sides are in proportion.

Recall that the sum of all 3 angles in a triangle total 180^{o}. This means that if you know that 2 angles in 2 different triangles are the same, then the 3rd angle must also be the same in both triangles. Therefore, the triangles are similar.

Notation: We use ABC to denote the measure of the angle formed by the line segments AB and BC. We use |AB| to denote the distance from point A to point B.

Proposition: Let ABC and DEF be two triangles. If ABC = DEF and ACB = DFE, then ABC and

|AC| |AB| |BC| ---- = ---- = ---- (see diagram below) |DF| |DE| |EF|

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*The Pythagorean Theorem:*

Similarity of triangles provides a method for proving one of the most important theorems in mathematics. A *right triangle* is one that contains a right angle - that is, a angle of 90^{o}. The side of the triangle opposite the right angle is called the *hypotenuse*; the remaining sides are called the *legs*.

*Theorem (Pythagoras)*: Let ABC be a right triangle with right angle at A. Let a and b be the lengths of the legs and let c be the length of the hypotenuse of ABC. Then c^{2} = a^{2} + b^{2}. That is, "The square of the hypotenuse is equal to the sum of the squares of the legs".

*Proof*: The diagram below shows ABC with the right angle at A at the top of the picture and BC as the base.

Construct the line perpendicular to BC that passes through the point A. Let D be the point of intersection of this line with BC. Since ABC and ACB are both less than a right angle, D lies between B and C as shown above.

Claim: The triangles ABC and DBA are similar.

This follows by the proposition stated earlier since the angle at B belongs to both triangles and BAC and BDA are both right angles. Likewise, the triangles ABC and DAC are similar.

Let d = |AD|, e = |BD|, and f = |DC| as shown above.

ABC similar to DBA ---> c/a = a/e ---> ce = a

^{2}

ABC similar to DAC ---> c/b = b/f ---> cf = b

^{2}

But e + f = c. So, a

^{2}+ b

^{2}= ce + cf = c(e + f) = c

^{2}.

Thus, c

^{2}= a

^{2}+ b

^{2}.

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*Distance between arbitrary points in the plane:*

We will now use the Pythagorean Theorem to create a formula for the distance between two points in terms of their coordinates.

Let P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) be two points in the plane. We can create a right triangle, by drawing a verticle line from Q and a horizontal line from P, as shown below.

The square of the bottom side is (x_{2} - x_{1})^{2} and the square of the verticle side is (y_{2} - y_{1})^{2}. If d denotes the distance between the two points P and Q, then from the Pythagorean Theorem: d^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}.

Taking square roots, we get the formula for the distance between the points:

_{2}- x

_{1})

^{2}+ (y

_{2}- y

_{1})

^{2}].

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