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, <BDE=<BCA (alternate angles). Hence, according to AA postulate, triangles ABC, and BDE are similar. But we know that the ratio of the length of corresponding sides are the same. If we mark this ratio with constant k, then we get

Thus, using the standard notation for sides of the triangle, we get that BD=a/k, DE=b/k, and EB=c/k. But, as said before BC=a, and AB=c, so we can express the length of AE, and DC in terms of k, and c and a respectively. Knowing that we have that DC=a(k-1)/k, and AE=c(k-1)/k.

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 <AHI=<EBD, and <HAI=<BED. Also, we have that GFAH is a parallelogram so GF=HA, and from before we know that GF=BE=c/k. Thus, according to ASA postulate those two triangles are congruent. From this we get that HI=a/k, and AI=b/k.

Since, CF=IA=b/k, and AC=b we get that IF=b(k-2)/k.

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 <BAC=<MIC, and <ABC=<IMC. Thus, according to AA postulate, triangles ABC, and IMC are similar. The fact that the triangle ABC, and FGC are similar was already proven. Hence, we have that triangles IMC, and FGC are similar as well. Knowing that we can look at the ratio of the lengths of corresponding sides. From that we get,

.

Since , we get that MC=GC(k-1).

Thus, MC=a(k-1)/k. From that we get,

But, we have shown before that

Since MG=DG, we can conclude that M=D. Thus, we have proven that the path Barney walks has a finite existence. From here we also see that Barney reaches the wall 5 times before returning to the initial point.

Q.E.D..