Final Assignment Part B


Exploring Altitude Ratios of an Acute Triangle

By: Sydney Roberts




For this final assignment, I chose to investigate the following conjecture: Given any acute triangle ABC, construct the Orthocenter H. Let points D, E, and F be the feet of the perpendiculars from A, B, and C respectfully. Prove the following:

To start this proof, itís helpful to look at a picture of a triangle with the different points labeled.


In order to see how these ratios hold, it is helpful to investigate the areas of triangles with the same base. To start this, consider the triangles with the base AC in the picture above. Then we want to find the area of these two triangles, and . Then we want to compare the ratios of these two areas.


Then we can see that

Next, consider the two triangles that share the base of AB and compare the ratio of their areas.


And lastly, consider the triangles that share the base of BC and compare the ratios of their area also.


Now that we see how these three different ratios can be written differently, we can use our area ratios to add together our original ratios because now they all have a common denominator.

However, now consider what the area summation would be for the numerator of the previous equation.


From this image, itís clear to see that

And hence, as desired.

Now we want to prove the second conjecture:

Conveniently, we can use our first conjecture in order to prove our second. Notice that the denominators in the two conjectures are the same. Through some simple line segment manipulation, we can change the numerators of the second conjecture as follows

We can use these manipulations and see that


Notice at this point that the summation within the parentheses is the same summation we proved was equivalent to 1 in the first conjecture. Hence, as desired.


Would these proofs hold if the triangle were obtuse?



If triangle ABC were obtuse, as shown above, then the orthocenter would no longer lie inside the triangle. Therefore, the methods we used before would not follow exactly. However, we could use the same methods and show that

This would work because and and and we can also see that .

Using these ratios, we could then apply the same method as above. Interestingly, we could also argue that this holds since by constructing , which is an acute triangle, the orthocenter would then be B, and then it follows that this conjecture would hold just as it did before with an arbitrary acute triangle.

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