**Part A**

**A. **
How do we prove that Barney will end up
where he started at? First of all, let's think about what we know about
triangles and some of their properties. Take a look at this picture.

In this picture, Barney starts at point D and goes to point E, walking parallel to segment AB. We know that triangle ABC is similar to triangle EDC because of the angle-angle-angle (AAA) theorem. Therefore, by definition of similar triangles, BC/DC = AC/EC. Let's call this number that they both equal "x". Alright, so Barney started at D and went to E, and now he moves parallel to segment BC.

Again, triangle AFE is similar to triangle ABC because of the AAA Theorem, so AB/FB = AC/EC = BC/DC = x. Now, I'm going to skip ahead a few steps. . .

Now, if we had went into detail about the steps in between, we would have proven that BA/IA = CA/HA = CB/GB = AB/FB = AC/EC = BC/DC = x. Now, a line drawn parallel to AC from point I to segment BC will intersect BC at a given point. Let's call this point J.

We know that BC/JC = BA/BI = . . . = x. However, we also know that BC/DC = x. So J = D!

One can see that Barney touches the triangle six times, twice on each side. This works for any point inside the triangle. (Obviously if Barney started at one of the vertices he would actually just walk around the perimeter of the triangle instead of inside of it.) An unusual case is when Barney starts at the midpoint of one of the sides. In this case, Barney only touches each side of the triangle once. (We even know the value of "x" in this case: 2! This is because the length of a whole side divided by the length of half of the same side equals 2.)

Now let's look at points on the exterior of triangle ABC.

Let's say that Barney starts at point D and moves to point E, walking parallel to AB. Do you notice anything? Triangles ABC and DCE are similar because of AAA Theorem! (The angles are equal because of alternate interior angles theorem and the fact that angle ACB = angle DCE.) So once again, let's say that BE/BC = AD/AC = x. And once again, let's skip. . .

At this point, we would have proven that AI/AB = HC/BC = GC/AC = FB/AB = BE/BC = AD/AC = x. Now, let's say Barney goes from point I parallel to BC and intersects line AD at a point J. So AJ/AC = AI/AB = . . . = x. But we already know that AD/AC = x. So J = D again!