Algebraic proof.

 If we look at the triangles ABC, and BDE we can see that both of them share the same angle at vertex B. Also, since DE and CA are parallel,

 If we draw a parallel to BC through E, then we get a parallelogram DCFE. Thus, DE=CF=b/k. Again, since DCFE is a parallelogram, DC=EF=a(k-1)/k. Having EF we got a new triangle, triangle AEF. It can easily be shown that triangles ABC, and AEF are similar. Having that in mind we have From this we get that , and that .

 From that again continuing Barney's walk we get point G on BC. Triangles BDE, and GCF are congruent because of ASA postulate (alternate angles). Thus, BD=GC=a/k, and DE=GF=c/k. Since BC=a, we get that DG=a-2a/k. Thus, we get that DG=a(k-2)/k.

 Constructing a parallel to AC through G, Barney reaches the wall at point H. By doing that we get a parallelogram GFAH. Thus, GF=HA=c/k. Again, we can conclude that triangles ABC, and HBG are similar because of AA postulate (alternate angles, and share the sam vertex and angle B). Also, since triangles ABC, and EBD are similar, we have that triangles HBG, and EBD are similar as well. Hence, From this we can get that HB=c(k-1)/k. Also, having in mind that BE=HA=c/k. we get that EH=c(k-2)/k.

 By drawing a line parallel to BC through point H we got point I on AC. By doing that we got a triangle AHI. Since HI is parallel to BC, then we have that

 Drawing a parallel to AB through, we get point M on BC. Have in mind that we are trying to prove that the path is finite. That is why we pick another point, in my case point M, If show that D=M, then we have proven our conjecture. What can we say about triangles ABC, and IMC? Well, they share the same angle at vertex C, and we said that IM is parallel to AB, so we have that

Go back.