Final Assignment

By

Brooke Norman

We
will first consider any triangle ABC.
Then we will select a point P inside the triangle and draw lines AP, BP,
and CP extended to their intersections with the opposite sides in points D, E,
and F respectively.

Now
lets explore (AF) (BD) (EC) and (FB) (DC) (EA) for various triangles and
various locations of P.

Click here to manipulate a GSP
sketch for different locations of P.

Did
you notice that as you moved P, (AF) (BD) (EC) = (FB) (DC) EA)?

Why
does this happen? Lets prove it!

First
we will construct parallel lines to produce similar triangles. Line HG is parallel to segment BC.

You
may notice that there appears to be several similar triangles. Lets compare two, AEH and CEB.

You can see that EH/EB = AE/EC = AH/BC. This proves that Triangle AHE is
similar to CEB.

You can also prove other triangles are similar, such
as Triangles FAG and FBC, and Triangles AHP and DBP.

After looking at all the segments, we find that:

AE/EC=AH/BC

BF/FA=CB/AG

AG/CD=AP/PD

AH/DB=AP/DP

From AH/CD=AP/PD and AG/DB=AP/DP you get AH/CD=AG/DB

Therefore you now know CD/DB=AH/AG

by multiplying the first, second and last equations,
we get:

(AE/EC)(BF/FA)(CD/DB) = (AH/CB)(CB/AG)(AG/AH) = (AH)(CB)(AG)/(CB)(AG)(AH) = 1

This proves that the ratio of the sides equals 1.

Does this proof hold true even when P is outside of
the triangle? Click here to explore it on your
own.

When P is inside triangle ABC, the ratio of the areas
of triangle ABC and triangle DEF are always greater than or equal to 4. When does it equal to 4?

The ratio equals exactly 4 when P is located at the
centroid of the triangle.

This concludes my exploration of the final
assignment. Hope you enjoyed it.

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6680 Homepage.