Thomas Earl Ricks
Consider any triangle ABC, like the one below.
Now choose any point P inside triangle ABC and draw lines AP, BP, and CP extended to their intersections with the opposite sides in points D, E, and F, like so:
Explore (AF)(BD)(EC) and (FB)(DC)(EA) for various triangles and locations of P.
Upon doing this, one quickly notices that (AF)(BD)(EC) and (FB)(DC)(EA) are equal to each other for various triangles and locations of P, through exploring in GSP.
I.e., it appears that
And it appears to matter not whether the shape of the triangle, or even if P is outside the triangle, (or on it). The property still appears to hold.
Click here for a working GSP sketch.
This does not constitute a proof, however.
I have not yet found a proof, but will present below what I have discovered.
I begin by drawing parallels to AD through F and E to obtain line segments FK and EM.
Then by properties of similar triangles, I obtain the ratios:
Then I solve each for AD, obtaining:
Since both ratios equal AD, I can set each equal to the other, obtaining the equality
Which allows me to divide on one side, obtaining:
Similarly, we construct parallel lines to BE through F and D to obtain line segments FN and DL, like so:
Which allows us to form the ratios
And solving both for BE, and then setting each equal to each other, we obtain the ratio:
And similarly we draw parallels to FC to obtain ER and QD,
which give us the ratios, by similar triangles, of:
And solving for FC, setting each equal to each other, we obtain:
We thus have three fractions equal to 1, namely:
We set the first two equal to each other
And divide, obtaining
And then we set this new fraction equal to the last of the three, obtaining
And then we divide again to obtain
Then we rearrange the terms, obtaining
which gives us three pieces which multiplied together equal 1:
We note that
Therefore, we now that our fraction is
which is the same as
So that the remaining two pieces must also equal 1
One of the pieces is what we wish to show equals one, which means if I can prove that
Then I have proved that
But I do not know how to do this.
But experimenting on GSP makes me believe that indeed
Click here for a file to manipulate showing the conjecture that the above ratio is be true.
We also were asked to show that if P is inside the triangle ABC, the ratio of the areas of triangle ABC and triangle DEF is always greater than or equal to 4. When is it equal to 4?
The area ratio is equal to 4 when points D, E, and F lie on the midpoints of the sides of triangle ABC. In other words, the area is equal to 4 when triangle DEF is the medial triangle for triangle ABC, like so:
It is equal to 4 because the area of triangle of DEF is one fourth the area of triangle ABC. This is true because AF=BF, AE=EC, and BD=DC since F, E, and D are midpoints. FE=1/2 of BC, so FE=BD=DC. Similarly, ED=AF=FB and FD=AE=EC. Therefore, triangle DEF is congruent to triangles AFE, FBD, and EDC. Thus four congruent triangles have equal area, so triangle DEF is one-fourth of the area of triangle ABC
In all other cases, when P is inside triangle ABC, triangle DEF is smaller than it was when forming the medial triangle, therefore, the ratio is greater than 4.
I am not sure how to prove that the largest triangle DEF can be is when it is the medial triangle and in all other cases it is smaller. But this must be done to complete the proof.