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, it’s
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, it’s
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 
.
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.