Similiarity question

[Johnny5]

You are given two right triangles, one with a larger area than the other.

 

Each of the triangles has the angle [math] \theta [/math] in it.

 

The legs of the smaller right triangle are a, b; with a>b.

The legs of the larger right triangle are A,B; with A>B.

 

I am trying to prove that:

 

[math] \frac{a}{b} = \frac{A}{B} [/math]

 

I just can't find a simple way to prove it. Anyone know how?

 

Thanks

[Callipygous]

do you mean an actual geometric proof? like those annoying things they made me do in 8th grade? or do you mean a logical proof? like just how to work through the facts you have to get to the result a/b=A/B?

 

i can help with the latter but not the former. i dont know the names of all those properties and crap they make you learn in geometry any more : P

[Johnny5]

do you mean an actual geometric proof? like those annoying things they made me do in 8th grade? or do you mean a logical proof? like just how to work through the facts you have to get to the result a/b=A/B?

 

i can help with the latter but not the former. i dont know the names of all those properties and crap they make you learn in geometry any more : P

 

Here is exactly what I want...

 

A logical proof' date=' but you can draw whatever you want to assist in the explanation.

 

I started out with one right triangle.

 

Then i picked a point at random on the hypotenuse.

 

Then, i constructed a perpendicular to one of the legs.

 

So then I had a right triangle inside a right triangle.

 

I then labled the legs a,b,A,B.

 

Now, I am just lookin for a proof that a/b=A/B.

 

If you prove that aB=Ab, then this will suffice for me.

 

aB is the area of a particular rectangle in the figure i started out with, and Ab is the area of a different rectangle in the figure i started out with.

 

But this drawing just led me in circles.

 

Then i started off with the assumption that

 

[math'] \exists r \in \mathbb{R} [ ar=A] [/math]

 

So r is some real number. So if the smaller triangle has a base of 10 feet, and the larger triangle has base of 100 feet, then r=10.

 

I am having trouble finding a proof for it.

 

The shorter and simpler the proof is the better.

 

I intend to use this as a lemma in a more complicated theorem.

[Callipygous]

if you know that they share one angle and they are both right triangles then you know that the other angle is also the same. they are definately similar triangles. i think the laws of similar triangles pretty much go straight from that to the kind of thing you are looking for. that is, if you know two triangles are similar then you also know that the ratio of a to A equals the ratio of b to B.

 

an actual proof is hard to get. the best way i can think of is to just assign some numbers to the sides and angles.

 

 

(my brain just totally died on me, sorry if this isnt what your looking for : P)

[Johnny5]

if you know that they share one angle and they are both right triangles then you know that the other angle is also the same. they are definately similar triangles. i think the laws of similar triangles pretty much go straight from that to the kind of thing you are looking for. that is' date=' if you know two triangles are similar then you also know that the ratio of a to A equals the ratio of b to B.

 

an actual proof is hard to get. the best way i can think of is to just assign some numbers to the sides and angles.

 

 

(my brain just totally died on me, sorry if this isnt what your looking for : P)[/quote']

 

Yes, they have all three angles the same, so they are definitely similiar.

 

Do you mean use analytic geometry, and do it by brute force?

[Callipygous]

sure.

 

pick numbers for the hypotenus(sp?) and angle of each, from there find the the lengths of the sides, devide them, and find out that they are in fact the same ratio.

[Johnny5]

sure.

 

pick numbers for the hypotenus(sp?) and angle of each' date=' from there find the the lengths of the sides, devide them, and find out that they are in fact the same ratio.[/quote']

 

Ok, if I cant find anything simpler, that's the way it's going to be forever.

 

Thanks

[uncool]

Would you accept proof through law of cosines?

-Uncool-

[Johnny5]

Would you accept proof through law of cosines?

-Uncool-

 

The law of cosines is more complex than what is trying to be proven. But I would enjoy seeing' date=' first a proof the law of cosines, and then subsequent to that a proof of this.

 

I have something I worked out, just this morning.

 

 

Begin with the definition of the area of a square:

 

[b']Definition: [/b] The area of a square is defined to be the product of its side with itself. That is:

 

[math] A_{square} \equiv S \cdot S = S^2 [/math]

 

Where S denotes the length of a side.

 

 

Theorem:

 

To prove: The area of a rectangle is equal to the product of its length, times its width. That is:

 

[math] A_{rectangle} \equiv l \cdot w [/math]

 

Where l denotes the length of the rectangle, and w denotes the width.

 

Solution:

 

Pick an arbitrary point at random, on one of the sides of a square, and at that point, construct the perpendicular, so that it passes through two sides of the given square.

 

The perpendicular line divides each side into two parts, one of length x, and the other of length y. And the whole side is equal to the sum of its parts, that is:

 

[math] S = x + y [/math]

 

Let is be the case that x<y.

 

Now construct a circle with radius x, using the corner of the given square as center.

 

At the point where the circle cuts this side of the given square, construct a line which is perpendicular to this side, and also perpendicular to the side opposite of it.

 

At this point, the original square is now equal to the sum of four smaller figures, two of which are squares, the remaining two shall be called rectangles.

 

We now wish to deduce the formula for the area of an arbitrary rectangle.

 

Denote the area of arbitrary rectangle as:

 

[math] A_{rectangle} [/math]

 

Now, the area of the whole square is equal to the sum of its parts. One of the parts is a square with side length x, and so has an area given by:

 

[math] A_1 = x^2 [/math]

 

And the truth of this, follows from the definition of the area of a square.

 

The other square has an area of:

 

[math] A_2 = y^2 [/math]

 

Again, this follows from the definition of the area of a square.

 

The remaining two figures, are what we have called 'rectangles', and they have equal areas, we are just trying to find a formula for the area of an arbitrary rectangle.

 

So, since the whole is equal to the sum of its parts, we have:

 

[math] S^2 = x^2 +y^2 + A_{rectangle} + A_{rectangle} [/math]

 

From which it follows that:

 

[math] S^2 = x^2 +y^2 + 2A_{rectangle} [/math]

 

 

But S = x+y, therefore S² is equal to (x+y)²

 

Now, using the transitive property of equality, it follows that:

 

[math] (x+y)^2 = x^2 +y^2 + 2A_{rectangle} [/math]

 

Now, if we make use of the fact following fact, which can be proven using algebra, we can finally find the formula for an arbitrary rectangle. Here it is:

 

[math] (x+y)^2 = x^2 +y^2+2xy [/math]

 

Now, using the transitive property of equality, we have:

 

[math] x^2 +y^2+2xy = x^2 +y^2 + 2A_{rectangle} [/math]

 

Now, subtract (x^2 +y^2) from each side of the statement above, to obtain the following statement:

 

[math] 2xy = 2A_{rectangle} [/math]

 

This statement has the same truth value of the previous statement. Since we know the previous statement is true, we know the one above is true as well.

 

Now, divide each side of the statement above by the number two, to obtain the following statement, which is also true:

 

[math] xy = A_{rectangle} [/math]

 

And there it is.

 

That is the formula for an arbitrary rectangle, with width x, and length y.

 

And the reason it is the formula for the area of an arbitrary rectangle, is because you could have started off with a square of any area imaginable, and still you get this formula, using deduction.

 

Now, from here, we can figure out the formula for the area of an arbitrary triangle. First, let us start off with a special case of the area of a right triangle, and then we shall generalize our knowledge, to that of any triangle.

 

I am going call this next theorem a Lemma, because it is a special case of a more powerful theorem, which will be proven immediately afterwards. Then the lemma no longer has to be remembered, so long as we remember the more powerful theorem, which immediately follows the lemma.

 

Lemma:

 

The area of a right triangle, is equal to one half of the product of its base times its height. That is:

 

[math] A_{right triangle} = \frac{b \cdot h}{2} [/math]

 

 

Let us start out with the arbitrary rectangle from the previous theorem. Its width was x, and its length was y.

 

Construct the diagonal of this rectangle. In order to do this, you choose one of the corners, and then draw a straight line to the corner of the rectangle which is furthest away from the point you chose. After that is done, that straight line is called a diagonal of the rectangle. Let this have been done.

 

You will now notice that the rectangle is now composed of two right triangles, of equal area. So that we can write this:

 

[math] A_{rectangle} = A_{right triangle} + A_{right triangle} = 2A_{right triangle} [/math]

 

But we already know the formula for the area of a rectangle, we found it in the previous theorem, so that using the transitive property of equality, we can now write the following formula, which is also true:

 

[math] x \cdot y = 2A_{right triangle} [/math]

 

Now, divide both sides of the statement above, by the number 2, to obtain the following statement, which has the same truth value as the previous one:

 

[math] \frac{x \cdot y}{2} = A_{right triangle} [/math]

 

And that is it, we are done, we have found the formula for the area of an arbitrary right triangle. Calling x the height, and y the base, we can write the formula as follows:

 

[math] A_{right triangle} = \frac{b \cdot h}{2} [/math]

 

 

And now to the more general case of an arbitrary right triangle. We now wish to prove that the area of any triangle, is equal to one half the product of its base, times its height. So we wish to prove the following theorem, which is more powerful than the previous Lemma:

 

Theorem:

 

The area of any triangle is one half the product of its base, times its height.

 

Suppose you are given an arbitrary triangle. Take the triangle and rotate it, so that its longest side, which shall be called the base of the triangle, is at at the bottom, now the height of this triangle is the length of a straight line which is perpendicular to the base, and passes through the apex of the triangle.

 

You will notice that after construction of this line, which is now the height h of the arbitrary triangle, the arbitrary triangle is now the union of two smaller triangles, both of which happen to be right triangles. And we just found the formula for the area of an arbitrary right triangle, in the previous Lemma.

 

Let us denote the length of the base of this triangle by the letter B. Let us denote the next smaller side, by the letter C, and let us denote the remaining side by the letter A.

 

Now, the straight line with length h, divides the base of the triangle into two parts, which i will just call b1, and b2. Thus, we the following statement is true:

 

[math] B = b_1 + b_2 [/math]

 

Because the whole equals the sum of its parts.

 

Let it be the case that b1 is less than b2.

 

Therefore, using the previous lemma, the area of the smaller of these two right triangles is given by:

 

[math] A_1 = \frac{b_1 \cdot h}{2} [/math]

 

And the area of the larger of these two right triangles is given by:

 

[math] A_2 = \frac{b_2 \cdot h}{2} [/math]

 

Therefore, the area of the whole triangle is:

 

[math] A_{triangle} = A_1 + A_2 = \frac{b_1 \cdot h}{2} + \frac{b_2 \cdot h}{2} [/math]

 

Where I have just substituted one equivalent expression for another.

 

Now, using algebra, we can convince ourselves that the following statement is true:

 

[math] \frac{b_1 \cdot h}{2} + \frac{b_2 \cdot h}{2} = (b_1 + b_2) \frac{h}{2} [/math]

 

Therefore, using the transitive property of equality, we can write:

 

[math] A_{triangle} = (b_1 + b_2) \frac{h}{2} [/math]

 

But b1+b2 is just the length of the arbitrary triangle, which we denoted by B.

 

Therefore we finally have the formula for the area of an arbitrary triangle, which is:

 

[math] A_{triangle} = B \frac{h}{2} [/math]

 

In words, we say, the area of a triangle is one half the product of its base times its height.

 

At this point in the argument, we will now prove the most famous theorem in the history of mathematics... the Pythagorean theorem. It was proven over two thousand years ago, supposedly, by the Greek mathematician Pythagoras. Special cases of this theorem were known to Egyptian mathematicians long before Pythagoras, but legend has it that Pythagoras was the first mathematician to prove that the statement is true, for all right triangles.

 

Theorem:

 

The area of the square on the hypotenuse of a right triangle, is equal to the sum of the areas of the squares on the legs. That is:

 

[math] h^2 = x^2 + y^2 [/math]

 

Where h denotes the longest side of a right triangle, called they hypotenuse, and x is the length of one leg of the right triangle, and y is the length of the other leg of the triangle.

 

Suppose we are given an arbitrary square with side length S. By definition, the area of this square is:

 

[math] S^2 [/math]

 

Pick a point, at random, on one of the sides of the given square. It thus divides the side of the square into two parts, one with length x, and the other with length y, so that we have:

 

[math] S = x + y [/math]

 

Let it be the case that x<y.

 

Now, it just so happens that Euclid of Alexandria knew of an amazing theorem, which is here:

 

Euclid, proposition 2

 

Now Euclid eventually proves Pythagoras' theorem, it is theorem 47 in Euclid's first of 13 books.

 

But we are going to use a different proof, discovered after Euclid, which makes use of Euclid's second proposition. In fact, we have already used some of Euclid's other theorems, which appear before they Pythagorean theorem, even though it wasn't explicitely stated.

 

So we have a given square, of side length S, and we have picked a point on it at random, which divides the side into two parts, one of length x, and the other of length y, and we have stipulated that x<y.

 

Now, using Euclid's second proposition, place a straight line with length x, at an adjacent corner of the given square. Now, using that line as radius, and the corner as center, describe a circle.

 

The circle so described, will cut the adjacent side at a point, call it point A, and the distance from point A to the center of the circle just described, will have length x.

 

Now, join this point A, to point B, where point B is the original point we randomly chose.

 

We have now just constructed a right triangle, with one leg of length x, another leg of length y, and a hypotenuse of length h.

 

Repeat this procedure three more times, to obtain the following figure:

 

A simple proof of the Pythagorean theorem

 

In the figure there, z is the hypotenuse, which we have denoted by h.

 

At this point it is easy to finish the proof. Since the whole equals the sum of its parts, we can write the following true statement:

 

[math] (x+y)^2 = \frac{xy}{2}+\frac{xy}{2}+\frac{xy}{2}+\frac{xy}{2}+ h^2 [/math]

 

We can factor out the 1/2 using the distributive axiom of algebra, to obtain the following statement which must have the same truth value as the previous statement:

 

[math] (x+y)^2 = \frac{1}{2}(xy+xy+xy+xy) + h^2 [/math]

 

And now we can factor out xy, as we did 1/2 to obtain:

 

[math] (x+y)^2 = \frac{xy}{2}(1+1+1+1) + h^2 [/math]

 

And we all know that 1+1+1+1=4, therefore:

 

[math] (x+y)^2 = \frac{xy}{2}(4) + h^2 [/math]

 

And 4 divided by two equals 2, therefore:

 

[math] (x+y)^2 = 2xy + h^2 [/math]

 

Recall that:

 

[math] (x+y)^2 = x^2 +y^2 +2xy [/math]

 

So now, we can use the transitive property of equality to conclude that:

 

[math] x^2 +y^2 +2xy = 2xy + h^2 [/math]

 

Now, subtract 2xy from both sides of the equation above, to obtain the following statement, which must also be true:

 

[math] x^2 +y^2 = h^2 [/math]

 

And the previous statement is Pythagoras' statement.

 

And we could have started off with a square as large as we wanted, therefore, the previous statement is true for any arbitrary right triangle, and that statement is called the Pythagorean theorem, in honor of Pythagoras who first proved it was true for any right triangle not just a specific one.

 

Before moving on to the next theorem, a special case of the Pythagorean theorem was known to the ancient Egyptians. It is easy to convince yourself that:

 

[math] 3^2+4^2 = 5^2 [/math]

 

[math] 3 \cdot 3 = 9 [/math]

[math] 4 \cdot 4 = 16 [/math]

[math] 5 \cdot 5 = 25 [/math]

[math] 9 + 16 = 25 [/math]

 

Therefore, using the transitive property of equality it follows that:

 

[math] 3^2+4^2 = 5^2 [/math]

 

And the Egyptians knew of this numerical fact. But, recall Pythagoras' theorem.

 

We already know that Pythagoras' theorem is true for any right triangle, so it will be true for a right triangle, with one leg of length 3, and the other leg of length 4.

 

And its good to think of this as a real figure in what is now called three dimensional Euclidan space, or Euclidean three-space for short.

 

Let your unit of distance be the meter. The meter is the currently accepted international unit of distance. One meter is slightly longer than three feet, and three feet is equivalent to one yard. Yards, and feet are units of the English system of units, but the meter is the unit of length of the metric system. That system of units is being used all over Europe right now.

 

At any rate, here is what the Egyptians knew. They knew that if they had a string of length seven feet, and they bent it at a right angle, in order to form a right triangle with one leg having length 3 meters, and the other length having length 4 meters, they knew, without even bothering to measure, that the hypotenuse would have length 5 meters. Certainly you could measure it, but you didn't have to.

 

Another way they looked at it, was as a way to construct a right triangle, and this helped them when they built the pyramids, which was around 2800 BC. The first pyramids were just squares on top of squares, called ziggurats, one of them can be seen here:

 

Ziggurat

 

So one way to construct a right triangle, would be to take a string of length 12 meter, and tie it together to form a closed loop.

 

Now, if you tighten the string correctly, to form a triangle with sides 3 meters, four meters, and five meters, one of the angles inside the triangle, will now be a right angle, and any triangle with a right angle in it, is a right triangle. And we know this is so, because we know Pythagoras' theorem.

 

Most likely you will not be able to get the legs of your triangle to be precisely 3 meters and four meters, but your triangle will be very nearly, a right triangle, it will be an approximation to a right triangle.

 

Let us summarize what we have done thus far.

 

We started off with the definition of the area of a square:

 

For any square, the area of the square is equal to the product of its side length with itself. As a formula we can write this as follows:

 

[math] \forall a \in A[ A_{\text{square x}} = S^2 ] [/math]

 

Where A denotes the set of all squares, and only squares.

 

Translation 1: For any a, which is an element of the set of squares, the area of a is equal to "S squared."

 

Translation 2: For any square a, the area of a is equal to S²

 

Translation 3: The area of any square is equal to the product of the length of its side with itself.

 

 

The sentence above was formulated using first order logic. Now in grammar we learn that sentences have to have a noun, and a verb, but this isn't always the case with a sentence formulated using first order logic. But, if you want to try to find the 'noun' of the sentence above, I guess you could say that the 'noun' is the set of all squares, and only squares. That really isn't a noun, but it is the main subject matter of the sentence. But first order logic doesn't need to obey the laws of grammar, instead the main thing about the sentence is that it needs to be either true, or false. And any sentence which is either true, or false, but not both simultaneously, is called a statement.

 

Now using the definition of square, we went on to discover the formula for the area of a rectangle deductively. Here is what we found, expressed using first order logic:

 

[math] \forall b \in B[ A_{\text{rectangle b}} = l \cdot w] [/math]

 

Translation1: For any b an element of the set of rectangles, the area of b is equal to the product of the length of rectangle b, times the width of rectangle b.

 

Translation2: For any rectangle b, the area of b is equal to the length of b, times the width of b.

 

Translation3: The area of any rectangle, is equal to its length times its width.

 

After deducing the formula for the area of an arbitrary rectangle, we went on and deduced the formula for the area of an arbitrary square. We can now formulate what we learned using first order logic, as follows:

 

[math] \forall c \in C [ A_{\text{triangle c}} = \frac{b \cdot h}{2} ] [/math]

 

Translation1: For any c an element of the set of triangles, the area of triangle c is equal to one half its base times its height.

 

Translation2: For any triangle c, the area of c is equal to one half its base times its height.

 

Translation3: The area of any triangle is equal to one half its base times its height.

 

And after this, we immediately proved the most famous theorem in the history of mathematics, the Pythagorean theorem. We can express the Pythagorean theorem using first order logic, as follows:

 

[math] \forall d \in D [ h^2 = x^2 + y^2] [/math]

 

Translation1: For any d an element of the set of right triangles, h squared = x squared plus y squared, where h is the hypotenuse of the right triangle, and x is the length of one leg, and y is the length of the other leg.

 

Translation2: For any right triangle d, h squared is equal to x squared plus y squared, where x is the length of one leg, and y is the length of the other leg.

 

Translation 3: In any right triangle, the square on the hypotenuse is equal to the sum of the squares on the legs.

 

Now in any right triangle, the three sides have different lengths. But this is not the case with an equilateral triangle. In any equilateral triangle, all three sides of the triangle must have exactly the same length. Furthermore, no figure composed of only straight lines, can have less sides than a triangle. A triangle is the simplest of all such closed figures.

 

Euclid expressed this fact by saying that, "two straight lines cannot enclose a space." This fact is self evident. Two straight lines either form an angle, or, are in a common infinitely long straight line.

 

Now, we used Euclid's second proposition in order to prove the Pythagorean theorem, but Euclid's second proposition required that his first proposition be previously proven. And his first proposition was the construction of an equilateral triangle. Later in book one, Euclid constructs a square. Now, it may occur to you, that we can keep increasing the number of sides of such figures. Figures, composed of only straight lines, in a common plane, which 'close.' The equilateral triangle is the simplest of these, the square the next, and then the regular pentagon, a regular hexagon, then a heptagon, all of whose sides have equal lengths, then an octogon, then a nonogon, then a decagon, then an 11 sided figure all of whose sides have the same length, and then a dodecagon, and so on.

 

All such figures are called regular polygons. The word 'regular' indicates that the length of any side is the same as the length of any other side.

 

Now, think about a circle. A circle is a figure that closes, but it isn't composed of any straight lines. It's smooth. It is a curved line, not a straight line, and Euclid took as postulate, that given any point as center and any straight line as radius, a circle could be made.

 

Now begin thinking about regular polygons, with n sides.

 

If n=2, there is no polygon formed, and you just have a bent line, an angle. Actually, the proper terminology is rectilinear angle, also called rectilineal angle. This is in contrast to a curvilinear angle. The difference between the two kinds of angles, is that one must have straight lines as sides, not so for the other kind.

 

If n=1, then you just have a straight line, with no angle.

 

But, if n=3, then you have the simplest of all polygons, a triangle. And if the three sides have the same length, then its called an equilateral triangle.

 

If n=4 its called a square.

 

Now, as n increases, the regular polygon looks more and more like a circle.

 

If we keep thinking in this manner, we can eventually deduce the formula for the area of a circle.

 

We already found formulas for the area of an arbitrary triangle, and the area of an arbitrary rectangle, starting with the definition of the area of a square.

 

Now, the best idea for the next theorem, is to construct a formula for the area of any regular polygon, in terms of the number of its sides n.

 

Then, we can let n approach infinity, to finally obtain the formula for the area of a circle. This was first deduced by Archimedes, the formula he got was:

 

[math] A_{\text circle} = \pi R^2 [/math]

 

Now, in order to work our way steadily towards this formula, let us introduce the definition of an apothem.

 

Here is the definition of apothem.

 

Here is a discussion about Archimedes' proofArea of a circle

 

Notice that it was proved by Carl Louis Ferdinand von Lindemann(1852-1939), that this is impossible, using only the moves allowed by Euclid.

 

If you read the article, you will see that von Lindemann was the first human being to prove that pi is a transcendental number.

 

What this means, is that pi is not the root of any algebraic equation with rational coefficients.

 

Let us set aside the theory of equations for now, and just focus on obtaining a formula for the area of a circle ourselves, using the idea of a limit.