The sum of angles in a polygon; more on reflections

Here is a demonstration (not a proof) that the sum of interior angles in a particular triangle is 180 degrees.

Draw any triangle on a piece of paper, using a ruler; cut it out, making sure the sides are straight.
Choose one side of the triangle as the base. If your triangle has an obtuse angle, use the side opposite the obtuse angle as the base. Fold the triangle on a line parallel to the base so that the top vertex lies on the base.
Now fold the triangle so that the right vertex touches the top vertex (in its folded down position). The two parts of the right side of the triangle should line up, and the right part of the bottom side should be folded along itself.
Fold in the left vertex in the same way. You should see the three interior angles of the triangle meeting at a point, forming an 180 degree angle with no overlapping.

Theorem: The sum of interior angles in any triangle is 180 degrees.

One of the extension problems is to make the demonstration just described into a formal proof. You will need the material on reflections farther down in this page.

Definition: The reflection of a point P across a line L is defined as follows. Construct the perpendicular to L through P. Call F the point where the perpendicular intersects L (the foot of the perpendicular). The reflection of P is at distance PF along the perpendicular on the other side of L. The reflection of P across L is denoted refL(P), or sometimes P'.

Reflection Theorem. Reflections preserve (a) distance and (b) angle.
That is, we are considering reflection across some specific line L. Then
  (a) the distance between any two points A to B equals the distance from the reflection of A to the reflection of B and
  (b) the measure of any angle AOB is the same as the measure of its reflection.

Proof of (a): Let A' and B' be the reflections of A and B, and F and G the feet of the perpendiculars to L, respectively.
Step 1: Show that triangle AFG is congruent to triangle A'FG. Side AF is congruent to side A'F, since A' is the reflection of A. Angle AFG is congruent to angle A'FG, because they are both right angles. Side FG is the same segment as side FG. Therefore, by the SAS triangle congruence theorem, triangle AFG is congruent to triangle A'FG.
Step 2: Show that triangle AGB is congruent to triangle A'GB'. Because triangle AFG is congruent to triangle A'FG, side AG is congruent to side A'G and angle AGF is congruent to angle A'GF. Angle BGF is congruent to angle B'GF because they are both right angles. Subtracting angle AGF from angle BGF, and angle A'GF from angle B'GF, we get that angle BGA is congruent to angle B'GA'. Side BG is congruent to side B'G, since B' is the reflection of B. Therefore, using the SAS triangle congruence theorem again, triangle BGA is congruent to triangle B'GA'.
Since corresponding parts of congruent triangles are congruent, AB is congruent to A'B'.

Proof of (b): Form triangle AOB from angle AOB by drawing segment AB. Let A', O', and B' be the reflections of A, O, and B, respectively. By part (a), AO=A'O', OB=O'B', and BA=B'A'. Using the SSS triangle congruence theorem, triangle AOB is congruent to triangle A'O'B'. So all corresponding angles are equal; in particular angle AOB is congruent to angle A'O'B'.

Tessellations by polygons

Recall that a tessellation of the plane by polygons is a collection of polygons that cover the palne without gaps or overlaps. The polygons are not necessarily congruent, and they do not necessarily make a repeating pattern. Here are two special types of tessellations.

Definition: In a tessellation where all the polygons are regular, the type of a vertex is a list of all the polygons around the vertex in a cyclic order. Usually a regular n-gon is just abbreviated by the number of sides n. So the type of each vertex in a checkerboard is 4.4.4.4. The type of each vertex in the equilateral triangle tessellation in the first lesson is 3.3.3.3.3.3.

Theorem: In any tessellation of the plane by polygons, the sum of the angles meeting at a vertex is 360 degrees.

Group problems

1. Find a formula for the sum of the interior angles in an n-gon (a polygon with n sides, where n is a positive whole number).
2. Find a formula for the measure of one angle in a regular n-gon. (All angles in a regular n-gon are equal.)
3. There is an edge-to-edge tessellation by regular polygons where every vertex has the same type, and some of the polygons are regular octagons.
1. What are the other polygons? (Hint: Use your formulas for angles in regular n-gons.)
2. Draw the tessellation.
3. Prove that this is the only tessellation with these rules that uses octagons.

Individual problems

1. The Reflection Theorem claims to be true for any points A and B.
1. Does the proof given work if A and B are on opposite sides of L? If not, adapt it so that it works in this case.
2. Does the proof work if one or both points are on L? If not, adapt it so that it works in this case.
2. Suppose you have a tessellation of the plane by polygons which is not edge-to-edge. What can you say about the angles at such a vertex?
3. Any triangle can be used to tessellate the plane (copies of only that triangle). Explain how to do this.
4. Find all edge-to-edge tessellation by regular polygons where every vertex has the same type, and the polygons are equilateral triangles and squares.
5. (Exension) Prove that you have found all the tessellations described in the previous problem.
6. (Extension) Make the demonstration of the 180 degrees theorem into a proof. Use any propositions, etc., from Book I of Euclid, and the Reflection Theorem above.