Audrey V. Simmons
Problem: Consider any triangle ABC. Select 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.
Explore (AF)(BD)(ED) and (FB)(DC)(EA) for various triangles and various locations of P.
To begin with we will pick various locations of point P to see what affect it has on the relationship of the segments of the sides of the triangle. Since some special features result with the centroid, we will pick the centroid as
When point P is the centroid, triangle ABC is divided into six triangles with equal area. Points F, E, and D are the midpoints of ABC and point P is the point of concurrency. Click here to see point P as the centroid.
The equality of the areas of these six triangles is true only when point P is the centroid
Not only are the areas equal, but if we multiply the length of the following sides together we end up with the equality
(AF)(BD)(ED) = (FB)(DC)(EA).
However, the question becomes, “Does this relationship hold true for all points P inside a given triangle or only when P is the centroid?”
By linking to the given triangle ABC we can move point P to various places within the triangle to see that this equality seems to be true for all possible P inside triangle ABC. Click here to link.
Is this relationship true only when point P is inside the triangle?
Click here to see.
Notice that points D,E,F leave the triangle through the vertices and not through the sides of triangle ABC.
We can prove that triangle BCA is similar to triangle CEG. Both triangles share angle ACB. By drawing a line through E and parallel to AB, we can show similarity by using the fact that corresponding angles are congruent.
Since all three angles are congruent, the triangles are similar.
Therefore the sides are proportional.
, , ,
By drawing a line through point F and parallel to BC (click here to see construction) we again have similar triangles with AFE and ABC. The proportions of the corresponding sides are as follows:
, , ,
By drawing an additional line through point D and parallel to AC, we have similar triangles in BDF and BCA. The same proof is done here as above. The following sides are proportional:
, , ,
Let’s see if we can make some connections with all these facts.
We know that
Take the reciprocal of the second proportion.
If we multiply them together we have:
That gives us:
I know there is a relationship of similar triangles that will give me the proportions that I am looking for, but I can not seem to find it. Parallel lines can also be drawn in a second direction from each of points D,E, and F which gives us other relationships, but I do not know how to prove the final equality.
I may be at a dead end. I have been told that Ceva’s theorem has a proof of this, but I have not been able to prove this on my own. I am also aware that part of this is on our class page, but I can not recall it on my own.
Looking back at triangle ABC and arbitrary point P inside the triangle, we can see through the GSP construction that the ratio of the areas of triangle ABC and triangle DEF is always greater than or equal to 4.
Click here to move point P to different locations inside the triangle and watch the value of the ratio.
The ratio is equal to four only when D,E,and F are the midpoints of triangle ABC and point P is concurrent with the centroid of the triangle. We then have a medial triangle formed by the points. Click here to see a construction.