Triangle ratios
Bennett Haselton, 2006/10/05
Suppose you are given a triangle ABC and an arbitrary point inside the triangle labeled
D. Draw lines through AD, BD and CD that meet the opposite sides in points labeled E, F,
and G, respectively:
The straight lines ADE, BDF and CDG are called "cevians", after the Italian mathematician
Giovanni Ceva, who studied them.
(The lines AGB, BEC and CFA are called "sides" after the American mathematician
Bart Side.)
Many math problems give you a triangle of the form above, give you the ratios that certain
sides or certain cevians are divided into, and ask you to compute the other ratios. For example,
suppose the lengths of the lines in the diagram above are given by this diagram:
How would you compute the unknown length DE?
There are six possible ratios present in a problem like that (one for each side, and one
for each cevian), and the formulae given here rely on a simple fact: If you know any two of the
ratios, you can compute the other four.
In the following diagrams, we select three of the possible sides/cevian ratios, and
then show the equation that relates the three values. Using this equation, if you have two of
the three values, you can compute the third one.
Ceva's Theorem
If you know the ratios that the sides are divided into, you can use Ceva's Theorem:
(Ironically, Ceva's Theorem is the only relationship listed here that doesn't actually
involve the cevians.)
Proof
The key to Ceva's Theorem is realizing that in the figure below:
(It's possible, but not easy, to "visualize" why this statement is true. We skip
the step of proving it formally.)
Similarly for the other sides, we have:
and it follows that if you multiply them together, all terms on the top and bottom cancel,
leaving 1.
Relationship between the three cevian ratios
The ratios that the cevians are divided into, are interrelated by the formula:
Proof
In this figure:
Similarly, we have:
So,
|
|
+ |
|
+ |
|
= |
| Area(ADB) + Area(BDC) + Area (CDA) |
|
| Area(ABC) |
|
But of course the sum of the areas of the three smaller triangles
equals the area of ABC, so the fraction reduces to 1.
Menelaus' theorem
This theorem describes the relationship between the ratios on two of the
sides, and the ratio on a cevian from the end of one side to a point
on the other side:
Menelaus' theorem states:
(Note: The theorem is usually not given in this format, but the other
formats are equivalent, and this format is the one that is most useful in
determining ratios of cevian and side segments in triangles.)
Proof
From this diagram:
we see a relationship between Area(BDC) and Area(ADC):
Then, using the segments along the side AFC:
we can show
the relationship between Area(ADC) and Area(CDF):
So, combining these two relationships, we get the ratio
of the areas of triangles BDC and FDC, but as this diagram shows,
that ratio is also given by the segments of the line BDF:
So:
Relationship between two cevians and the side between them
This theorem describes the relationship between the ratio on one of the triangle's
sides, and the ratios on the two cevians that start at either end of that side:
Proof
We can see from this diagram:
And similarly, from this diagram:
So multiplying these ratios gives us the ratio between Area(ADB) and Area(ADC).
But as this diagram shows:
So,
Combining these results
By using the results above, you can start with any two of the six ratios and find the
rest. Going back to the problem posted at the beginning:
We know the ratios for two of the three sides, so we can compute the ratio for the
third side: BE:EC = 15:14. So we can label those segments 15k and 14k:
Now switch focus to a different set of 3 sides, ones that are arranged in a pattern
for which we can use Menelaus' Theorem:
Labelling the unknown length x, by Menelaus' Theorem we have:
So x = 84/29.