Do you understand the title?
The foot of an altitude
Consider with vertices shown below:
is the “foot” of the altitude . The coordinates of can be obtained by solving the equations of and simultaneously. This way we have:
Due to the conspicuousness of the terms and in the above expressions, both and can be recast into a different form.
Let . PROVE that the coordinates of the foot of the altitude in the above diagram can be re-written as: .
The above equivalent representations are more convenient for computations. To see why the first one is true, simply divide every term in the equation
by . (Note that, in view of the original diagram, , and so division by doesn’t result in any absurdity.)
Similarly, by dividing every term in the expression for by , 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.
has vertices at . Find the coordinates of the foot of the perpendicular from vertex .
From the diagram above, the foot of the perpendicular from vertex appears to be the point . Let’s confirm this using our formula. To this end, let . The starting ingredient we need is . Next, the “common denominator” in the simplified expressions for and is . Putting , this becomes .
So it’s verified that the altitude’s foot is at . This can also be checked by the regular, preferred method (that is, finding the equation of side and the equation of the altitude from , then finding their point of intersection).
has vertices at . PROVE that the length of the altitude from vertex is:
Recall that the coordinates of the foot of the altitude from vertex are given by:
The distance from to is, by the distance formula:
Note that both and have a common factor of . Squaring, adding, simplifying, then taking square root gives
has vertices at . PROVE that the area of can be given by:
The area of a triangle follows easily:
A negative sign in the final result can be ignored.
Find the area of with vertices at .
This is a right triangle whose legs are on the coordinate axes, so we expect to get square units as the area.
To use the area formula, let . Then the area is:
Find the area of with vertices at .
Let . Then:
The required area is 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.
Suppose that we have an isosceles in which :
If , then . (In particular, if the base is “flat” — in the sense that — then is just the average of and .)
To see this, simply use the fact that 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.)
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 is , then
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).
The following are equivalent:
It is easier to prove that by expanding and simplifying the terms. It is also easy to prove that ; just carefully retrace your steps backwards.
The above equivalence always holds regardless of what the ‘s and ‘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.
Let have vertices at . Then the foot of the altitude from divides the base in the ratio In other words, a negative sign in the final result can be ignored.
PROVE the Pythagorean theorem: In a right triangle :
Recall that the foot of the perpendicular from to divides in the ratio
In the present case, the foot coincides with , and so the left member of this ratio is zero; that is, . Based on the equivalent conditions in Example 8 above, this is the same as: So, .
PROVE the Pythagorean theorem: In a right triangle
The -coordinate of the foot of the altitude from to side is given by
In the present case, this is precisely , and so
In view of the equivalent conditions in Example 8 above, we conclude that or that . 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?
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.
- has vertices at . Determine the coordinates of the foot of the altitude from .
- Consider the diagram at the beginning of this post:
PROVE that the foot of the altitude from divides the base in the ratio In other words, PROVE that 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 with vertices at be such that . PROVE that , where .
[Use the fact that the foot of the altitude from to divides in the ratio ]
- In any non-right triangle with vertices , PROVE that and .
- PROVE that [And so we have an equivalent expression for the area of a triangle.]
- Let be the vertices of a quadrilateral. PROVE that the area of the quadrilateral can be given by .
[You can divide the quadrilateral into two triangles, then use the formula for the area of a triangle. Then .]