4 Lengths, areas and proportions

We now turn to a brief discussion of area. Rather than carefully developing this theory, we shall begin with some ``obvious'' facts, such as

(i)

The area of a rectangle which has side lengths $a$ and $b$ is $ab$.

(ii)

Congruent figures have equal area

(iii)

The area of a region obtained as a union of two non-overlapping regions is the sum of the areas of these regions.

4.1 Right triangles

A right triangle is one which contains a right angle. The two sides emanating from the right angle are called the legs of the triangle; the third side, opposite the right angle is the hypoteneuse.

Proposition 4.1   If the legs of a right triangle have lengths $a$ and $b$, then the area of the triangle is $\frac{1}{2} ab$.

Exercise 4.1   Prove this proposition by constructing an appropriate rectangle.

Proposition 4.2   In $\triangle ABC$, if $D$ is the endpoint of the altitude from $A$, then the area of $\triangle ABC$ is equal to $\frac{1}{2}\vert AD\vert\vert BC\vert$.

Exercise 4.2   Prove this proposition.

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.

figure=pix/pythag.eps,width=.2                          figure=pix/pythag-euler.eps,width=.2

Theorem 4.3 (The Pythagorean Theorem)   Let the lengths of the legs of a right triangle be $a$ and $b$, and let the length of the hypoteneuse be $c$. Then $a^2+b^2=c^2$.

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.

Proof. The proof is essentially the figure shown. We start with a right triangle with legs of length $a$ and $b$ and hypotenuse of length $c$. Now we construct three more copies of this triangle, arranging them to construct a square of side length $a+b$ as in the figure. This is indeed a square because all four sides are of length $a+b$, and each angle of the outer quadrilateral is a right angle.

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 $c$. 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 $\pi$. Two of these are the acute angles in a right triangle, and so the third must be of measure $\pi/2$.

The area of the outer square is $(a+b)^2$, the area of each of the four triangles is $\frac{1}{2} ab$, and the area of the inner square is $c^2$. Thus, we have

$$(a+b)^2=4(\frac{1}{2} ab)+c^2$$

From this, we readily see that $a^2+b^2=c^2$. $\qedsymbol$

Exercise 4.3   Locate a different proof of the Pythagorean Theorem, and explain it in your own words.

4.2 Similar triangles

We say that two triangles $ABC$ and $A'B'C'$ are similar if $m\angle A=m\angle A'$, $m\angle B=m\angle B'$ and $m\angle C=m\angle C'$.

Since the sum of the measures of a triangle is constant, if we are given two triangles, $\triangle ABC$ and $\triangle A'B'C'$, and we know that $m\angle A=m\angle A'$, and $m\angle B=m\angle B'$, then the triangles are similar.

If the triangles are similar, we write $\triangle ABC\sim\triangle A'B'C'$.

Proposition 4.4   Let $D$ be some point on the side $AB$ of $\triangle ABC$, and let $k$ be the line through $D$ parallel to $BC$. Let $E$ be the point where $k$ crosses $AC$. Then $\triangle ABC\sim\triangle ADE$.

Exercise 4.4   Prove this proposition.

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.

Lemma 4.5 (Ratios for right triangles)   Let $\triangle ABC\sim\triangle A'B'C'$, where $C$ and $C'$ are right angles. Then

$$\frac{\vert AB\vert}{\vert A'B'\vert}=\frac{\vert BC\vert}{\vert B'C'\vert}=\frac{\vert CA\vert}{\vert C'A'\vert}.$$

Proof. Without loss of generality, we can assume that $\vert AB\vert>\vert A'B'\vert$, and so we start with $\triangle ABC$, and construct another triangle which is congruent to $\triangle A'B'C'$ inside it.

Let $k_1$ be a line parallel to $BC$ that meets $AB$ at the point $D$ where $\vert AD\vert=\vert A'B'\vert$. Denote the point where $k_1$ meets $AC$ by $E$. Let $k_2$ be the line through $A$ parallel to $BC$, and let $m_2$ be the line perpendicular to $AC$ at $B$. Then, since $AC$ and $m_2$ are both perpendicular to $BC$, they are parallel. Let $F$ be the point of intersection of $m_2$ with $k_2$. Then $AFBC$ is a rectangle. Also, let $G$ be the point of intersection of $m_2$ and $k_1$. Finally, let $m_1$ be the line perpendicular to $k_1$ and $k_2$ through $D$; denote the point of intersection of $m_1$ with $BC$ by $H$ and denote the point of intersection of $m_1$ with $k_2$ by $I$.

Since the lines $BC$, $k_1$ and $k_2$ are all parallel, and the lines $AC$, $m_1$ and $m_2$ are all parallel, $\vert BH\vert=\vert GD\vert=\vert FI\vert$, $\vert HC\vert=\vert DE\vert=\vert IA\vert$, $\vert BG\vert=\vert HD\vert=\vert CE\vert$, and $\vert GF\vert=\vert DI\vert=\vert EA\vert$. We give names to these quantities; we set $a=\vert BH\vert=\vert GD\vert=\vert FI\vert$, $b=\vert HC\vert=\vert DE\vert=\vert IA\vert$, $c=\vert BG\vert=\vert HD\vert=\vert CE\vert$ and $d=\vert GF\vert=\vert IE\vert=\vert EA\vert$. We also set $e=\vert BD\vert$ and $f=\vert DA\vert$.

The line $AB$ divides the rectangle $AFBC$ into two congruent triangles. Since the areas of $\triangle BGD$ and $\triangle BHD$ are equal, and the areas of $\triangle IDA$ and $\triangle EAD$ are equal, we obtain that the areas of the two smaller rectangles are equal; that is, $ad=bc$.

Since we drew the line $EG$ parallel to $BC$, we know that $\triangle AED\sim\triangle ACB$. So, for this case, our theorem, which we want to prove, says that

$$\frac{d}{c+d}=\frac{b}{a+b}=\frac{f}{e+f}.$$

To see that the first of these inequalities is true, notice that it holds when $d(a+b) = b(c+d)$ (by cross multiplying). But this holds since we have already established that $ad=bc$.

We check that the second equality is true by using the first equality together with the Pythagorean theorem. That is, we write $f^2=b^2+d^2$ and $(e+f)^2=(a+b)^2+(c+d)^2$, and then, as above, cross-multiply and use the two facts that we have already proven; namely that $ad=bc$, and that

$$\frac{d^2}{(c+d)^2}=\frac{b^2}{(a+b)^2}.$$

Now since $\triangle ABC\sim\triangle A'B'C'$, we have $m\angle A=m\angle A'$ and $m\angle B=m\angle B'$. Since $k_1$ is parallel to $BC$, $m\angle B = m\angle EDA$ (using alternate interior angles and vertical angles). Since $\vert AD\vert=\vert A'B'\vert$ by construction, $\triangle A'B'C' \cong \triangle ADE$ via ASA. Using the above, we conclude that

$$\frac{\vert AB\vert}{\vert A'B'\vert}=\frac{\vert BC\vert}{\vert B'C'\vert}=\frac{\vert CA\vert}{\vert C'A'\vert}$$

as desired. $\qedsymbol$

Now that we have shown the desired property for right triangles, we can use that result to show it for arbitrary similar triangles.

Theorem 4.6 (Ratios for triangles)   Let $\triangle ABC\sim\triangle A'B'C'$. Then

$$\frac{\vert AB\vert}{\vert A'B'\vert}=\frac{\vert BC\vert}{\vert B'C'\vert}=\frac{\vert CA\vert}{\vert C'A'\vert}.$$

Proof. As above, we can assume without loss of generality that $\vert AB\vert>\vert A'B'\vert$. Find the point $D$ on $AB$ so that $\vert AD\vert=\vert A'B'\vert$, and draw the line $DE$ parallel to $BC$. As above, using alternate interior angles, we see that $m\angle ADE=m\angle ABC$ and $m\angle AED=m\angle ACB$. Hence $\triangle ADE\cong\triangle A'B'C'$, and so it suffices to show that

$$\frac{\vert AB\vert}{\vert AD\vert}=\frac{\vert BC\vert}{\vert DE\vert}=\frac{\vert CA\vert}{\vert EA\vert}.$$

Let $k=AF$ be the altitude from $A$, and let $G$ be the point of intersection of $k$ with the line $DE$. Then $ADF$ and $ADG$ are similar right triangles, and $ACF$ and $AEG$ are similar right triangles. Hence the proportion theorem for right triangles yields

$$\frac{\vert AD\vert}{\vert AB\vert}=\frac{\vert AG\vert}{\vert AF\vert}=\frac {\vert AE\vert}{\vert AC\vert}.$$

That $\frac{\vert DE\vert}{\vert BC\vert}=\frac{\vert AG\vert}{\vert AF\vert}$ follows easily from the Lemma above. $\qedsymbol$

Exercise 4.5   Prove that the lines joining the midpoints of the sides of a triangle divide the triangle into four congruent triangles.

4.3 Basic trigonometric functions

Let $\triangle ABC$ be such that $\angle C$ is a right angle. Let $a=\vert BC\vert$, $b=\vert AC\vert$ and $c=\vert AB\vert$. Also, let $\theta=m\angle A$. Then, by the proportion theorem for right triangles, the ratios $a/b$, $a/c$, and $b/c$ all depend only on the measure of the angle $\theta$.

In what follows, we will deliberately confuse an angle with its measure; i.e., we will not distinguish between $\angle A$ and $m\angle A$. It should always be clear from the context which is meant.

For $0 < \theta < \pi/2$, we define the trigonometric functions as ratios of lengths in right triangles:

$\sin\theta=a/c$

$\cos\theta=b/c$

$\tan\theta=a/b=\sin\theta/\cos\theta$

$\cot\theta=b/a=1/\tan\theta=\cos\theta/\sin\theta$

$\sec\theta=c/b=1/\cos\theta$

$\csc\theta=c/a=1/\sin\theta$

We note that it is almost immediate from the definition that

$$\sin(\pi/2-\theta)=\cos\theta$$   and$$\qquad \cos(\pi/2-\theta)=\sin\theta.$$

In order to extend the definitions to all values of $\theta$, first we set

$$\sin 0 = 0 \qquad\hbox{and}\qquad \sin\frac{\pi}{2} = 1$$

which is reasonable since the sine is small for small angles, and close to 1 for angles close to right angles. Now to define obtuse angles,

$$\sin(\pi-\alpha)=\sin(\alpha) \qquad\hbox{for}\qquad 0\le\alpha \le\frac{\pi}{2}$$

and to get negative values, we let

$$\sin(-\theta)=-\sin(\theta).$$

Finally, since an angle of $2\pi$ represents a full turn, we say that $\sin(2\pi + \alpha) = \sin\alpha$. Putting these together defines the sine for all values.

We can then use the fact that $\sin(\pi/2-\theta)=\cos\theta$ and $\tan\theta = \sin\theta/ \cos\theta$ 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 $\tan\frac{\pi}{2}$.

Exercise 4.6   Prove that, for any number $\theta$,

$$\sin^2\theta+\cos^2\theta = 1.$$

Theorem 4.7 (Law of Sines)   Let $ABC$ be a triangle, with sides of length $a=\vert BC\vert$, $b=\vert AC\vert$ and $c=\vert AB\vert$. Then

$$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}.$$

Proof. We first assume that $C$ is a right angle, and, that $c=1$. Then $\sin A=a$ and $\sin B=b$, so the result holds in this case. Using the theorem of proportions for right triangles, our result holds for all right triangles.

Now let $ABC$ be an arbitrary triangle, where $B$ and $C$ are acute angles (i.e., their measures are less than $\pi/2$), and let $AD$ be the altitude from $A$. Then $\sin B=\frac{\vert AD\vert}{c}$, and $\sin C=\frac{\vert AD\vert}{b}$. Hence

$$\frac{\sin B}{b}=\frac{\vert AD\vert}{bc}=\frac{\sin C}{c}.$$

                

We next take up the case that angle $B$ is an obtuse angle. We again drop the altitude $AD$ from $A$ to the line $BC$, but now this altitude lies outside $\triangle ABC$. Note that $B$ lies between $D$ and $C$. As above, using our known facts about right triangles, we obtain, $\vert AD\vert=\vert AC\vert\sin C=\vert AB\vert\sin B$, from which the desired result follows.

If angle $C$ is obtuse, repeat the above argument, exchanging $B$ and $C$.

The other equality follows by looking at the altitude from $B$ and/or $C$. $\qedsymbol$

Theorem 4.8 (Law of Cosines)   Let $ABC$ be a triangle, where none of the angles are right angles, with sides of length $a=\vert BC\vert$, $b=\vert AC\vert$ and $c=\vert AB\vert$. Then

$$c^2=a^2+b^2-2ab\cos C.$$

Proof. As above, we need to worry about the possibility that $\angle C$ is obtuse. We first take up the case that it is acute. It could be that either $\angle A$ or $\angle B$ is obtuse, we note that our hypotheses, and the formula we wish to prove, remain unchanged if we interchange $A$ and $B$; we assume that $\angle B$ is not obtuse. It then follows that if $AD$ is the altitute from $A$, then $D$ lies between $A$ and $B$. Let $d=\vert AD\vert$. Let $a_1=\vert BD\vert$, and let $a_2=\vert DC\vert$. Then, by the Pythagorean theorem, $c^2=a_1^2+d^2$. Again by the Pythagorean theorem, $d=b^2-a_2^2$. Substituting, we obtain

$$c^2=a_1^2+d^2=a_1^2 + b^2- a_2^2=(a_1+a_2)^2+b^2-2a_1a_2-2a_2^2=a^2+b^2-2aa_2=a^2+b^2-2ab\cos C,$$

where, in the last equality, we have used the fact that $\cos C=a_2/b$.

We next take up the case that $\angle C$ is obtuse. In this case, the endpoint of the altitude $AD$ is such that $C$ lies between $B$ and $D$. Similar to the above, we set $a_1=\vert CD\vert$. Then the Pythorean theorem yields the following.

$$(a+a_1)^2+d^2=c^2,\quad d^2+a_1^2=b^2.$$

Substituting, we obtain,

$$c^2=(a+a_1)^2+b^2-a_1^2=a^2+b^2+2aa_1.$$

We note that $\cos(\pi-C)=a_1/b$. Hence, upon setting $\cos C=-\cos(\pi-C)$, we obtain,

$$c^2=a^2+b^2+-2ab\cos(\pi-C)=a^2+b^2-2ab\cos C.$$

$\qedsymbol$

4.4 Other important trigonometric formulae

We will not prove the addition formulae for sines and cosines; however, since they are important, we will state them.

Theorem 4.9  

$$\sin(\alpha+\beta)=\sin \alpha\cos\beta+\sin\beta\cos\alpha.$$

Theorem 4.10  

$$\cos(\alpha +\beta)=\cos\alpha \cos\beta -\sin\alpha \sin\beta.$$

One can easily prove these theorems by constructing $\triangle ABC$, where $m\angle A=\alpha +\beta$, and the altitude from $A$ divides this triangle into two triangles, $\triangle ABD$ and $\triangle ADC$, in such a way that $m\angle BAD=\alpha$ and $m\angle DAC=\beta$. 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.

Exercise 4.7   This exercise tells you how to compute the other three pieces of information about a triangle if you are given SAS.

(a)

Assume we are given the lengths $a$ and $b$ of triangle $ABC$, and we are given $\angle C$ (that is, we are given $\sin C$ and/or $\cos C$). Explain in words how we would find the length of the other side, and the other two angles. (Hint: Use the above theorems.)

(b)

Find explicit formulas for these three pieces of information in explicit terms; that is, find a formula for $c$, and find formulas for the other two angles, in terms of $a$, $b$ and either $\sin C$ or $\cos C$.

Exercise 4.8   Repeat the above for the congruence relation ASA. That is, if you are given the sines and or cosines of two of the angles, and given the length of the included side, find the sine or cosine of the other angle and find the lengths of the other two sides.

Exercise 4.9   Repeat the above exercise for AAS.

Exercise 4.10   Repeat the above exercise for SSS.