Altitude foots Pythagorean theorem – High School Geometry

Do you understand the title?

The foot of an altitude

Consider $\triangle ABC$ with vertices $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$ shown below:

$N(a,b)$ is the “foot” of the altitude $AN$ . The coordinates of $N$ can be obtained by solving the equations of $AN$ and $BC$ simultaneously. This way we have:

$\begin{equation*} \begin{split} a&=\frac{x_1(x_3-x_2)^2+x_3(y_3-y_2)^2+(y_1-y_3)(y_3-y_2)(x_3-x_2)}{(x_3-x_2)^2+(y_3-y_2)^2}\\ b&=\frac{y_1(y_3-y_2)^2+y_3(x_3-x_2)^2+(x_1-x_3)(x_3-x_2)(y_3-y_2)}{(x_3-x_2)^2+(y_3-y_2)^2} \end{split} \end{equation}$

Due to the conspicuousness of the terms $(x_3-x_2)$ and $(y_3-y_2)$ in the above expressions, both $a$ and $b$ can be recast into a different form.

Example 1

Let $m=\frac{y_3-y_2}{x_3-x_2}$ . PROVE that the coordinates of the foot $N(a,b)$ of the altitude in the above diagram can be re-written as: $a=\frac{x_1+m^2x_3+m(y_1-y_3)}{1+m^2}, \quad b=\frac{m^2y_1+y_3+m(x_1-x_3)}{1+m^2}$ .

The above equivalent representations are more convenient for computations. To see why the first one is true, simply divide every term in the equation

$a=\frac{x_1(x_3-x_2)^2+x_3(y_3-y_2)^2+(y_1-y_3)(y_3-y_2)(x_3-x_2)}{(x_3-x_2)^2+(y_3-y_2)^2}$

by $(x_3-x_2)^2$ . (Note that, in view of the original diagram, $x_3\neq x_2$ , and so division by $x_3-x_2$ doesn’t result in any absurdity.)

$\begin{equation*} \begin{split} a&=\frac{\frac{x_1(x_3-x_2)^2+x_3(y_3-y_2)^2+(y_1-y_3)(y_3-y_2)(x_3-x_2)}{(x_3-x_2)^2}}{\frac{(x_3-x_2)^2+(y_3-y_2)^2}{(x_3-x_2)^2}}\\ &=\frac{\frac{x_1(x_3-x_2)^2}{(x_3-x_2)^2}+\frac{x_3(y_3-y_2)^2}{(x_3-x_2)^2}+\frac{(y_1-y_3)(y_3-y_2)(x_3-x_2)}{(x_3-x_2)^2}}{\frac{(x_3-x_2)^2}{(x_3-x_2)^2}+\frac{(y_3-y_2)^2}{(x_3-x_2)^2}}\\ &=\frac{x_1+x_3\Big(\frac{y_3-y_2}{x_3-x_2}\Big)^2+(y_1-y_3)\Big(\frac{y_3-y_2}{x_3-x_2}\Big)\Big(\frac{x_3-x_2}{x_3-x_2}\Big)}{1+\Big(\frac{y_3-y_2}{x_3-x_2}\Big)^2}\\ &=\frac{x_1+x_3(m^2)+(y_1-y_3)m}{1+m^2}\quad\textrm{since}~m=\frac{y_3-y_2}{x_3-x_2}\\ &=\frac{x_1+m^2x_3+m(y_1-y_3)}{1+m^2} \end{split} \end{equation}$

Similarly, by dividing every term in the expression for $b$ by $(x_3-x_2)^2$ , we obtain the required simplified form.

From these simplified expressions, we see that once we find the slope of the side where the “foot” of the altitude lies, then the rest is easy.

Example 2

$\triangle PQR$ has vertices at $P(-3,1),~Q(3,-2),~R(0,2)$ . Find the coordinates of the foot of the perpendicular from vertex $R$ .

From the diagram above, the foot of the perpendicular from vertex $R$ appears to be the point $(-1,0)$ . Let’s confirm this using our formula. To this end, let $(x_1,y_1)=(0,2),~(x_2,y_2)=(-3,1),~(x_3,y_3)=(3,-2)$ . The starting ingredient we need is $m=\frac{y_3-y_2}{x_3-x_2}=\frac{-2-1}{3--3}=\frac{-3}{6}=-\frac{1}{2}$ . Next, the “common denominator” in the simplified expressions for $a$ and $b$ is $1+m^2$ . Putting $m=-\frac{1}{2}$ , this becomes $1+(-\frac{1}{2})^2=5/4$ .

$\begin{equation*} \begin{split} a&=\frac{x_1+m^2x_3+m(y_1-y_3)}{1+m^2}\\ &=\frac{0+(-\frac{1}{2})^2(3)+(-\frac{1}{2})(2--2)}{5/4}\\ &=\frac{0+(\frac{1}{4})(3)+(-\frac{1}{2})(4)}{5/4}\\ &=\frac{\frac{3}{4}-2}{5/4}\\ &=\frac{\frac{3}{4}-\frac{8}{4}}{5/4}\\ &=\frac{-5/4}{5/4}\\ &=-1\\ &\vdots\cdots\cdots\cdots\vdots\\ &\vdots\cdots\cdots\cdots\vdots\\ b&=\frac{m^2y_1+y_3+m(x_1-x_3)}{1+m^2}\\ &=\frac{(-\frac{1}{2})^2(2)+(-2)+(-\frac{1}{2})(0-3)}{5/4}\\ &=\frac{\frac{2}{4}-2+\frac{3}{2}}{5/4}\\ &=\frac{\frac{2}{4}-\frac{8}{4}+\frac{6}{4}}{5/4}\\ &=\frac{0}{5/4}\\ &=0 \end{split} \end{equation}$

So it’s verified that the altitude’s foot is at $(-1,0)$ . This can also be checked by the regular, preferred method (that is, finding the equation of side $PQ$ and the equation of the altitude from $R$ , then finding their point of intersection).

Example 3

$\triangle ABC$ has vertices at $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$ . PROVE that the length of the altitude from vertex $A$ is: $AN=\frac{y_1(x_3-x_2)+y_2(x_1-x_3)+y_3(x_2-x_1)}{\sqrt{(x_3-x_2)^2+(y_3-y_2)^2}}.$

Recall that the coordinates of the foot of the altitude from vertex $A(x_1,y_1)$ are given by:

The distance from $A(x_1,y_1)$ to $N(a,b)$ is, by the distance formula:

$AN=\sqrt{(a-x_1)^2+(b-y_1)^2}.$

Now

$\begin{equation*} \begin{split} a-x_1&=\frac{(y_3-y_2)\Big(y_1(x_3-x_2)+y_2(x_1-x_3)+y_3(x_2-x_1)\Big)}{(x_3-x_2)^2+(y_3-y_2)^2}\\ &\cdots\vdots\vdots\vdots\cdots\\ b-y_1&=\frac{(x_3-x_2)\Big(y_1(x_2-x_3)+y_2(x_3-x_1)+y_3(x_1-x_2)\Big)}{(x_3-x_2)^2+(y_3-y_2)^2}\\ &=-\frac{(x_3-x_2)\Big(y_1(x_3-x_2)+y_2(x_1-x_3)+y_3(x_2-x_1)\Big)}{(x_3-x_2)^2+(y_3-y_2)^2} \end{split} \end{equation}$

Note that both $a-x_1$ and $b-y_1$ have a common factor of $(y_1(x_3-x_2)+y_2(x_1-x_3)+y_3(x_2-x_1))$ . Squaring, adding, simplifying, then taking square root gives $AN=\frac{y_1(x_3-x_2)+y_2(x_1-x_3)+y_3(x_2-x_1)}{\sqrt{(x_3-x_2)^2+(y_3-y_2)^2}}.$

Example 4

$\triangle ABC$ has vertices at $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$ . PROVE that the area of $\triangle ABC$ can be given by: $\frac{y_1(x_3-x_2)+y_2(x_1-x_3)+y_3(x_2-x_1)}{2}$

The area of a triangle follows easily:

$\begin{equation*} \begin{split} \textrm{\textbf{AREA}}&=\frac{\textrm{base}\times\textrm{height}}{2}\\ &=\frac{BC\times AN}{2}\\ &=\frac{\sqrt{(x_3-x_2)^2+(y_3-y_2)^2}\times\Big(\frac{y_1(x_3-x_2)+y_2(x_1-x_3)+y_3(x_2-x_1)}{\sqrt{(x_3-x_2)^2+(y_3-y_2)^2}}\Big)}{2}\\ &=\frac{y_1(x_3-x_2)+y_2(x_1-x_3)+y_3(x_2-x_1)}{2} \end{split} \end{equation}$

A negative sign in the final result can be ignored.

Example 5

Find the area of $\triangle ABC$ with vertices at $A(0,0),~B(3,0),~C(0,4)$ .

This is a right triangle whose legs are on the coordinate axes, so we expect to get $\frac{3\times 4}{2}=6$ square units as the area.

To use the area formula, let $(x_1,y_1)=(0,0),~(x_2,y_2)=(3,0),~(x_3,y_3)=(0,4)$ . Then the area is:

$\begin{equation*} \begin{split} \textrm{\textbf{AREA}}&=\frac{0(0-3)+0(0-0)+4(3-0)}{2}\\ &=\frac{0+0+12}{2}\\ &=\frac{12}{2}\\ &=6\quad\textrm{square units} \end{split} \end{equation}$

Example 6

Find the area of $\triangle PQR$ with vertices at $P(-3,1),~Q(3,-2),~R(0,2)$ .

Let $(x_1,y_1)=(-3,1),~(x_2,y_2)=(0,2),~(x_3,y_3)=(3,-2)$ . Then:

$\begin{equation*} \begin{split} \textrm{\textbf{AREA}}&=\frac{y_1(x_3-x_2)+y_2(x_1-x_3)+y_3(x_2-x_1)}{2}\\ &=\frac{1(3-0)+2(-3-3)+(-2)(0--3)}{2}\\ &=\frac{1(3)+2(-6)-2(3)}{2}\\ &=\frac{3-12-6}{2}\\ &=\frac{-15}{2} \end{split} \end{equation}$

The required area is $\frac{15}{2}$ square units.

Altitude’s foot in an isosceles triangle

We take advantage of our general formula to prove a well-known property about isosceles triangles.

Equipment

Suppose that we have an isosceles $\triangle ABC$ in which $AB=AC$ :

If $m=\frac{y_3-y_2}{x_3-x_2}$ , then $x_1=\frac{x_2+x_3-m(2y_1-y_2-y_3)}{2}$ . (In particular, if the base is “flat” — in the sense that $y_2=y_3$ — then $x_1$ is just the average of $x_2$ and $x_3$ .)

To see this, simply use the fact that $AB=AC$ and the distance formula to derive an algebraic expression involving the vertex coordinates. This can then be put in the above form. (Another approach to this question is given in the exercises at the end.)

Example 7

PROVE that the foot of the altitude from the “apex” of an isosceles triangle bisects the base.

Please understand that there’s a simpler way to prove the above geometric property other than what we show below (see our comment at the end).

Let’s work with the diagram in the “Equipment” above. Suppose that the foot of the altitude from $A(x_1,y_1)$ is $N(a,b)$ , then

$\begin{equation*} \begin{split} a&=\frac{x_1+m^2x_3+m(y_1-y_3)}{1+m^2}\\ &=\frac{\frac{x_2+x_3-m(2y_1-y_2-y_3)}{2}+m^2x_3+m(y_1-y_3)}{1+m^2}\\ &=\frac{x_2+x_3-2my_1+my_2+my_3+2m^2x_3+2my_1-2my_3}{2(1+m^2)}\\ &=\frac{x_2+x_3+m(y_2-y_3)+2m^2x_3}{2(1+m^2)}\\ &=\frac{x_2+x_3-m^2(x_3-x_2)+2m^2x_3}{2(1+m^2)}\quad\textrm{since}~m=\frac{y_3-y_2}{x_3-x_2}\\ &=\frac{x_2+x_3-m^2x_3+m^2x_2+2m^2x_3}{2(1+m^2)}\\ &=\frac{x_2+m^2x_2+x_3+m^2x_3}{2(1+m^2)}\\ &=\frac{x_2(1+m^2)+x_3(1+m^2)}{2(1+m^2)}\\ &=\frac{(1+m^2)(x_2+x_3)}{2(1+m^2)}\\ &=\frac{x_2+x_3}{2}\\ &\vdots\cdots\cdots\cdots\vdots\\ &\vdots\cdots\cdots\cdots\vdots\\ \textrm{Similarly,}~b&=\frac{y_2+y_3}{2} \end{split} \end{equation}$

This proves that the altitude from the apex of an isosceles triangle bisects the base.

“Foot proof”: Pythagorean theorem

Now’s the main moment. We want to use the foot of the altitude to derive the Pythagorean theorem (this was what we meant by altitude “foots” — as in supplies or provides (a proof of) — the Pythagorean theorem, in the title).

Example 8

The following are equivalent:

$(x_1-x_2)(x_3-x_2)+(y_1-y_2)(y_3-y_2)=0$
$(x_1-x_3)^2+(y_1-y_3)^2=(x_1-x_2)^2+(y_1-y_2)^2+(x_3-x_2)^2+(y_3-y_2)^2$

It is easier to prove that $(2)\implies (1)$ by expanding and simplifying the terms. It is also easy to prove that $(1)\implies(2)$ ; just carefully retrace your steps backwards.

The above equivalence always holds regardless of what the $x$ ‘s and $y$ ‘s represent — so long as they’re real numbers. However, when these are the coordinates of the vertices of a triangle, they remind us that perpendicular line segments (can be connected to form a right triangle and so) imply Pythagorean theorem, and vice versa. Trivial fact, right? Almost quite.

Equipment

Let $\triangle ABC$ have vertices at $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$ . Then the foot $N$ of the altitude from $A$ divides the base $BC$ in the ratio $\Big((x_1-x_2)(x_3-x_2)+(y_1-y_2)(y_3-y_2)\Big):\Big((x_1-x_3)(x_3-x_2)+(y_1-y_3)(y_3-y_2)\Big).$ In other words, $\frac{BN}{CN}=\frac{(x_1-x_2)(x_3-x_2)+(y_1-y_2)(y_3-y_2)}{(x_1-x_3)(x_3-x_2)+(y_1-y_3)(y_3-y_2)};$ a negative sign in the final result can be ignored.

Example 9

PROVE the Pythagorean theorem: In a right triangle $ABC$ :

we have $AC^2=AB^2+BC^2.$

Recall that the foot of the perpendicular from $A$ to $BC$ divides $BC$ in the ratio

$\Big((x_1-x_2)(x_3-x_2)+(y_1-y_2)(y_3-y_2)\Big):\Big((x_1-x_3)(x_3-x_2)+(y_1-y_3)(y_3-y_2)\Big).$

In the present case, the foot coincides with $B$ , and so the left member of this ratio is zero; that is, $(x_1-x_2)(x_3-x_2)+(y_1-y_2)(y_3-y_2)=0$ . Based on the equivalent conditions in Example 8 above, this is the same as: $(x_1-x_3)^2+(y_1-y_3)^2=(x_1-x_2)^2+(y_1-y_2)^2+(x_3-x_2)^2+(y_3-y_2)^2.$ So, $AC^2=AB^2+BC^2$ .

Example 10

PROVE the Pythagorean theorem: In a right triangle $ABC$

we have $AC^2=AB^2+BC^2.$

The $x$ -coordinate of the foot of the altitude from $A(x_1,y_1)$ to side $BC$ is given by

$x=\frac{x_1(x_3-x_2)^2+x_3(y_3-y_2)^2+(y_1-y_3)(y_3-y_2)(x_3-x_2)}{(x_3-x_2)^2+(y_3-y_2)^2}.$

In the present case, this is precisely $x_2$ , and so

$\begin{equation*} \begin{split} \frac{x_1(x_3-x_2)^2+x_3(y_3-y_2)^2+(y_1-y_3)(y_3-y_2)(x_3-x_2)}{(x_3-x_2)^2+(y_3-y_2)^2}&=x_2\\ x_1(x_3-x_2)^2+x_3(y_3-y_2)^2+(y_1-y_3)(y_3-y_2)(x_3-x_2)&=x_2\Big((x_3-x_2)^2+(y_3-y_2)^2\Big)\\ (x_1-x_2)(x_3-x_2)^2+(x_3-x_2)(y_3-y_2)^2+(y_1-y_3)(y_3-y_2)(x_3-x_2)&=0\\ (x_1-x_2)(x_3-x_2)+(y_3-y_2)^2+(y_1-y_3)(y_3-y_2)&=0\quad\textrm{since}~x_3-x_2\neq 0\\ (x_1-x_2)(x_3-x_2)+(y_3-y_2)\Big(y_3-y_2+y_1-y_3\Big)&=0\\ (x_1-x_2)(x_3-x_2)+(y_3-y_2)(y_1-y_2)&=0\\ \end{split} \end{equation}$

In view of the equivalent conditions in Example 8 above, we conclude that $(x_1-x_3)^2+(y_1-y_3)^2=(x_1-x_2)^2+(y_1-y_2)^2+(x_3-x_2)^2+(y_3-y_2)^2,$ or that $AC^2=AB^2+BC^2$ . So we reobtain the Pythagorean theorem by considering a special case of the foot of an altitude.

Examples 9 and 10 illustrate a procedure we refer to as “foot proof”. Note, however, that it is not foolproof. In fact, can you spot the “bug” in this method?

Takeaway

It is often useful to approach a problem from two (or more) different perspectives, if possible. In this post (and in many of our posts), we choose a long, direct route, in a bid to keep to just the concept(s) under consideration.

When a problem is viewed from another perspective, it can happen that considerable simplification to the problem is obtained, as a lot of detail become abstracted away. Think about our proof of the fact that the altitude from the apex of an isosceles triangle bisects the base. This can be proved with a single line argument using “congruence”.

After solving a problem, we encourage you to always look for alternative — and possibly better — approaches.

Tasks

$\triangle ABC$ has vertices at $A(-3,-2),~B(3,1),~C(5,-1)$ . Determine the coordinates of the foot of the altitude from $B$ .
Consider the diagram at the beginning of this post:

PROVE that the foot of the altitude from $A$ divides the base in the ratio $[(x_1-x_2)(x_3-x_2)+(y_1-y_2)(y_3-y_2)]:[(x_1-x_3)(x_3-x_2)+(y_1-y_3)(y_3-y_2)].$ In other words, PROVE that $\frac{BN}{CN}=\frac{(x_1-x_2)(x_3-x_2)+(y_1-y_2)(y_3-y_2)}{(x_1-x_3)(x_3-x_2)+(y_1-y_3)(y_3-y_2)};$ a negative sign in the final result can be ignored.
[The result still applies in the case of an external altitude, though the diagram is different.]
Let $\triangle ABC$ with vertices at $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$ be such that $AB=AC$ . PROVE that $x_1=\frac{x_2+x_3-m(2y_1-y_2-y_3)}{2}$ , where $m=\frac{y_3-y_2}{x_3-x_2}$ .
[Use the fact that the foot of the altitude from $A$ to $BC$ divides $BC$ in the ratio $1:1$ ]
In any non-right triangle $ABC$ with vertices $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$ , PROVE that $(x_1-x_2)(x_3-x_2)+(y_1-y_2)(y_3-y_2)\neq 0$ and $(x_1-x_3)(x_3-x_2)+(y_1-y_3)(y_3-y_2)\neq 0$ .
PROVE that $y_1(x_3-x_2)+y_2(x_1-x_3)+y_3(x_2-x_1)=x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2).$ [And so we have an equivalent expression for the area of a triangle.]
Let $(x_1,y_1),~(x_2,y_2),~(x_3,y_3),~(x_4,y_4)$ be the vertices of a quadrilateral. PROVE that the area of the quadrilateral can be given by $\frac{(x_1-x_3)(y_2-y_4)+(x_2-x_4)(y_1-y_3)}{2}$ .
[You can divide the quadrilateral into two triangles, then use the formula for the area of a triangle. Then $\cdots$ .]