We now turn to a brief discussion of area. Rather than carefully developing this theory, we shall begin with some ``obvious'' facts, such as
The Pythagorean Theorem, relating the lengths of the legs of a right
triangle to the length of the hypotenuse, is very well-known; you've
probably seen it repeatedly since your elementary school days, and
know it quite well.
The figure on the left below is sometimes described as a proof of the
Pythagorean Theorem, but in fact it is much closer to a geometric
statement of the theorem. To turn it into a proof requires
constructing a number of additional lines, and a fairly complex
argument. On left is the figure for Euclid's proof, sometimes called
``the bride's chair''. We won't give the argument
here, but you might try to figure it out on your own. The idea is to
show that there are two pairs of congruent triangles, and the area of each
triangle in a pair is half of one of the smaller squares. The sum of
the areas in the pairs gives the area of the larger square.
There are many proofs of this theorem; we know of at least 40. Below we give one of the simpler ones. It is necessary to use the Parallel Axiom (Axiom 8) axiom, either implicitly or explicitly, in order to prove the Pythagorean Theorem; the theorem is false in both spherical and hyperbolic geometry, which have a different version of Axiom 8. Surprisingly, the Pythagorean Theorem is equivalent to this axiom; that is, the Parallel Axiom can be proven if we assume the Pythagorean Theorem first.
In addition, there is an inner quadrilateral formed whose sides are the
hypotenuse of each of the triangles. This quadrilateral is certainly a
rhombus, because each side is of length . But it is in fact a square,
because each of its angles are right: at each vertex there are three angles
which sum to an angle of measure
. Two of these are the acute angles in
a right triangle, and so the third must be of measure
.
The area of the outer square is , the area of each of the four
triangles is
, and the area of the inner square is
.
Thus, we have
From this, we readily see that
.
Since the sum of the measures of a triangle is constant, if we are given two
triangles,
and
, and we know that
, and
, then the triangles are similar.
If the triangles are similar, we write
.
We want to prove the very useful fact that the side lengths of similar
triangles are proportional. However, this is easier to do if we
first prove it for right triangles, and then apply this result to the
general case.
Let be a line parallel to
that meets
at the point
where
. Denote the point where
meets
by
. Let
be the line through
parallel to
, and let
be the line perpendicular to
at
. Then, since
and
are both perpendicular to
, they are parallel. Let
be the
point of intersection of
with
. Then
is a
rectangle. Also, let
be the point of intersection of
and
. Finally, let
be the line perpendicular to
and
through
; denote the point of intersection of
with
by
and
denote the point of intersection of
with
by
.
The line divides the rectangle
into two congruent triangles.
Since the areas of
and
are equal, and the
areas of
and
are equal, we obtain that the
areas of the two smaller rectangles are equal; that is,
.
Since we drew the line parallel to
, we know that
. So, for this case, our theorem, which we want to
prove, says that
To see that the first of these inequalities is true, notice that it
holds when
(by cross multiplying). But this holds
since we have already established that
.
We check that the second equality is true by using the first equality
together with the Pythagorean theorem. That is, we write
and
, and then, as above, cross-multiply and use the
two facts that we have already proven; namely that
, and that
Now since
, we have
and
. Since
is parallel to
,
(using alternate interior angles and
vertical angles). Since
by construction,
via ASA.
Using the above, we conclude that
Now that we have shown the desired property for right triangles, we can use that result to show it for arbitrary similar triangles.
Let be the altitude from
, and let
be the point of
intersection of
with the line
. Then
and
are
similar right triangles, and
and
are similar right
triangles. Hence the proportion theorem for right triangles yields
That
follows easily from the
Lemma above.
In what follows, we will deliberately confuse an angle with its
measure; i.e., we will not distinguish between and
. It should always be clear from the context which is meant.
We note that it is almost immediate from the definition that
In order to extend the definitions to all values of , first we set
We can then use the fact that
and
to extend the definitions of the other
functions to all real numbers (except those where we would need to divide by
0, such as for
.
Now let be an arbitrary triangle, where
and
are acute angles
(i.e., their measures are less than
),
and let
be the altitude from
. Then
, and
. Hence
If angle is obtuse, repeat the above argument, exchanging
and
.
The other equality follows by looking at the altitude from and/or
.
We next take up the case that is obtuse. In this case, the
endpoint of the altitude
is such that
lies between
and
. Similar to the above, we set
. Then the Pythorean
theorem yields the following.
One can easily prove these theorems by constructing
,
where
, and the altitude from
divides
this triangle into two triangles,
and
,
in such a way that
and
. Then,
using the definitions of the sine and cosine for right triangles, the
addition formula for the sine follows from the law of sines, and the
addition formula for the cosine follows from the Pythagorean theorem
and the law of cosines.
We can use the laws of sines and cosines to obtain information about
triangles.