by Rives Poe
For this assignment,
we will start by constructing a triangle ABC. Then 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. Below is such aconstruction.
Now, let's explore
(AF)(BF)(EC) and (FB)(DC)(EA) for various triangles and various
locations of P.
Click here to see and manipulate a GSP sketch for different
locations of P.
In this sketch
you should have noticed that (AF)*(BF)*(EC) = (FB)*(DC)*(EA).
So, we see that
they are equal, but we need to prove it!
Here goes nothing!
First we need to
construct parallel lines to produce similar triangles.
(Line HG is parallel
to segment BC. )
First we can see
that triangles EGA and ECB are similar:
= GE/EB = AE/CE.
Triangle FAH is
similar to triangle FBC.
= FA/FB = FH/FC
Triangles AGP and
DBP are similar:
Thus, AG/DB = GP/PB
And finally, triangle
HAP is similar to triangle CDP:
= AP/DP = HP/PC.
Now we need to
pull all of this information together to see if we can prove:
So, we need to
gather some information from above. We don't need all of the ratios,
just a few that are going to help us find the above ratio.
1. AG/BC =AE/CE
2.BC/HA = FB/FA
3. AH/DC= AG/DB**
**we can find use
this ratio because the ratios of triangle AGP and DBP are similar
to the ratios of triangles AHP and DCP.
So, when we multiply
the above ratios together we get:
some things cancel
out, so we are left with:
AG/DC = (AE/CE)(FB/FA)(AG/DB)
the left side by DB/AG, so that we now have on the left DB/DC.
Now, if we move
everything to the left we have:
And there is our
proof that the ratio of the sides equals 1!!!
to see a GSP
sketch of the triangle with lines, so that we can move P outside
of the triangle.
see a construction showing that when P is inside triangle ABC,
the ratio of the areas of triangle ABC and triangle DEF is always
greater than or equal to 4. The ratio is equal to 4 when P falls
on the centroid of ABC.
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