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:

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:

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: