Thomas Earl Ricks
Mathematics Education
Assignment # 8
Investigation 12
Orthocentric Ratios
For investigation 12, we construct any triangle ABC,
and then we construct the altitudes, creating points H
(the orthocenter) and D, E, and F, like so:
For ease of identification, we will only look at the
segments inside the triangle:
The first question is to prove:
To do this, we start with:
and multiply each part by a multiple of one, like so:
We then note that for each of the denominators, they
equal twice the area of triangle ABC:
since each denominator has an altitude of triangle ABC
and a base of triangle ABC. Thus
we can combine the three fractions with a common denominator like so:
and then we note that each piece on the top is twice a
certain triangle inside triangle ABC, namely:
This allows us to re-write the numerator in our
fraction to become:
and then after factoring out the common 2 on the
numerator, we get:
which 2 on the numerator reduces to one after division
with the 2 in the denominator, so we now get:
but then we notice that the three triangle areas on
the top are none other than the area of triangle ABC!
Therefore,
which allows us to combine the top numerator to just
the area of triangle ABC, which we also have in the denominator, so our
fraction equals 1:
Which means that our original ratio is 1:
Which completes our proof! Therefore,
For the second part of investigation 12, we want to
show that for the same triangle ABC and constructed perpendiculars,
the ratio:
We start with
And want to show this equals 2.
To do this, we note that we can convert each numerator
in the expression above into an equivalent form, like so:
this allows us to make the substitutions:
so that we can break each fraction up into two
fractions,
then we rearrange, like so:
bringing out the negative sign from the last three
fractions:
and then we factor out the negative sign from the last
three fractions:
and then observe that the first three fractions are each
equal to 1:
and then we remember in the first part of this webpage
that
which allows us to make the substitution for the last
three fractions, obtaining:
which means we have proved:
