**The Problem**

Barney is in the triangular room shown
here. He walks from a point on **Segment BC** parallel to **Segment**
**AC**. When he reaches **Segment AB**, he turns and walks
parallel to **Segment BC**. When he reaches **Segment BC**,
he turns and walks parallel to **Segment AB**. Prove that Barney
will eventually return to his starting point. How many times will
Barney reach a wall before returning to his starting point?

**The Solution**

The following figures chart Barney's route from the starting point,
**S.0**, to the bouncing points **S.1**, **S.2**, etc.
Note that segments containing two "**S**" points,
such as **S.1** and **S.2**, will be labeled as **Segment
S.1:S.2**

It appears that a path from **S.5**
parallel to **Segment AB** will pass through **S.0**. To
prove this, assume **Segment S.0:S.5** is not parallel to Segment
**AB**.

**Proof**

Consider the figure below:

From the description of Barney's path, three sets of parallel segments are present.

**Segments AC**,
**S.3:S.4**, and **S.0:S.1**

**Segments AB**
and **S.2:S.3**

**Segments BC**,**
S.1:S.2**, and **S.4:S.5**

From these parallel segements, parallelograms are formed. These parallelograms are:

**B:S.1:S.2:S.3**
(labeled in red)

**A:S.2:S.3:S.4**
(labeled in green)

From the parallel lines and parallelograms, it can be shown that the yellow triangles in the figure below are congruent.

**Angle A**
is congruent to **Angle B:S.1:S.0** because they are corresponding
angles formed by two parallel lines and a transversal. **Angle
A:S.4:S.5** and **Angle B** are congruent for the same reason.
**Segments A:S.4** and **B:S1** are congruent because they
are both congruent to **Segment S.2:S.3** (opposite sides of
a parallelogram).

Thus, the yellow triangles, **Triangle
A:S.4:S.5** and **Triangle S.1:A:S.0** are congruent by angle-side-angle.

From this congruence, **Segment S.4:S.5**
is congruent to **Segment B:S.0** (CPCTC).

Since these segments are congruent and
parallel (**Segment** **B:S.0** is part of **Segment BC**),
**Quadrilateral B:S.4:S.5:S.0** is a parallelogram. Thus, **Segment
S.5:S.0** is parallel to **Segment AB** and must be part
of Barney's path.

Thus Barney will return to his starting point after 5 bounces, or six trips across the interior of the triangle.

An exception to this statement occurs when Barney begins at the midpoint of a side of a triangle. In this case, Barney's route will trace the medial triangle and he will return to his starting point after only two bounces (three trips across the interior of the triangle).

**Extension**

What if Barney began at a point on **Line
BC**, but not on **Segement BC**? One such path is depicted
in the figure below.

Note that all of the sides of **Triangle
ABC** must be extended in order to chart Barney's path. Can
you prove that Barney's next move will take him back to **S.0**?
Click Here to access a Geometer's Sketchpad file containg
this sketch.