Final Assignment

by

Scott Burrell

A. Consider any triangle ABC. 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 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.

You should notice that (AP)(BD)(EC)=(FB)(DC)(EA)

B. Now we must prove why (AP)(BD)(EC)=(FB)(DC)(EA)

First we need to construct parallel lines to produce similar triangles. Below, Line HG is parallel to segment BC

We observe that several triangles appear to be similar. Below is a comparison of similar triangles EGA and ECB.

Therefore, AG/BC=GE/EB=AE/CE....triangle EGA is similar to triangle ECB.

Similar observations can be made by comparing triangles, FAH and FBC, AGP and DBP, and HAP and CDP. Click here to make the comparisons using a GSP script tool.

After making the comparisons and pulling everything together, you can conclude with the following:

AE/EC=AG/BC

BF/FA=CB/AH

AH/CD=AP/PD

AG/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

Now, by multiplying the first, second and last equations, we get:
(AE/EC)(BF/FA)(CD/DB) = (AG/CB)(CB/AH)(AH/AG) = (AG)(CB)(AH)/(CB)(AH)(AG) = 1

So we have proved that the ratio of the sides equals one.

Does this hold true even when P is outside of the triangle? Click here to explore further in GSP.

C. Click here to 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. Notice that the ratio is equal to 4 when P falls on the centroid of ABC.

This concludes my investigation of the final assignment. Several of my ideas came from Ceva's Theorem at www.cut-the-knot.org/Generalization/cevas.html. I would advise you to visit the site for more in depth information.

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