Final Assignment

# Semians and Similar Triangles

If we take any triangle ABC, and 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, we get a construction like the one below.

Now we want to explore the connection between the semians FB, DC, and EA; and AF, BD, and, EC.

We can see that even when we move P, the result is the same FB*DC*EA=AF*CE*BD.

If we want to prove the above conjecture, we begin by constructing a line parallel to BC through the point A, and then extend the lines AC and AB so they intersect our new line through A. Label the points of intersections H and I.

Now since we have parallel lines and intersecting lines we can conclude that the following triangles are similar.

Triangle BPD is similar to triangle IPA

Triangle HAP is similar to triangle CDP

Triangle AEI is similar to triangle CEB

And triangle HFA is similar to triangle FCB

Now we can conclude that the following ratios are true

since triangles BPD and IPA are similar. since triangles HAP and CDP are similar.

since triangles AEI and CEB are similar. since triangles HFA and FCB are similar.

So when we do a little manipulation of our ratios we see the following

so

and

so when we plug in AP from above, we get

also

so , so when we plug this into our equation above we get

and also so and we can put this into our equation, and get .

Now when we rearrange our equation we get

and our PD's and BC's cancel, so we have

We can also see a relationship between the triangle ABC and the new triangle formed by D, E, and F.

No matter where we move P to the ration of the area of triangle ABC to the area of triangle FED is greater than or equal to 4.

And when the ratio is equal to four, we notice that P is the centroid of triangle ABC.