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 180o. 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 DEF are similar. Moreover,


                          |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 90o. 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 c2 = a2 + b2. 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 = a2
ABC similar to DAC ---> c/b = b/f ---> cf = b2
But e + f = c. So, a2 + b2 = ce + cf = c(e + f) = c2.
Thus, c2 = a2 + b2.
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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(x1, y1) and Q(x2, y2) 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 (x2 - x1)2 and the square of the verticle side is (y2 - y1)2. If d denotes the distance between the two points P and Q, then from the Pythagorean Theorem: d2 = (x2 - x1)2 + (y2 - y1)2.
Taking square roots, we get the formula for the distance between the points:

d = sqrt[ (x2 - x1)2 + (y2 - y1)2].


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