Hexagon proofs

Yesterday, I was solving a math problem with my 13 year old son involving a hexagon. The sum of internal angles of an n-sided polygon = (n-2)*180°. For a hexagon, the number of sides n=6, the sum is (6-2)*180°=720°. I didn't like to use this formula and thought about a proof. Let's draw four triangles inside the hexagon and label the angles:

The 6 interior angles of the hexagon would be:

1. a1+a2+a3+a4
2. b1
3. c1+b2
4. c2+b3
5. c3+b4
6. c4

We know that the internal angles of a triangle, e.g. a1+b1+c1 is 180°. If we sum all the 6 internal angles of the hexagon, we see that they are equal to the sum of the internal angles of our 4 triangles. Therefore, the sum of the internal angles of a hexagon is 4*180°=720°.

Then I wanted to prove that the 6 triangles created by the diagonals of a regular hexagon are equilateral:

We can label the triangle internal angles x, y, z as follows:

 Using parallel lines, we can fill the internal angles of all 6 triangles:

Note that the sum of internal angles of the hexagon are (z+x)+(y+z)+(x+y)+(z+x)+(y+z)+(x+y) = 4*(x+y+z). Since x+y+z=180°, this is another way to prove that the sum of internal angles are 4*180° = 720°. Also note that this proof is only valid for a regular hexagon. For a non regular hexagon, the sides would not be parallel and we would not be able to assert the equality of angles of triangles. The first proof at the top is valid even for a non-regular hexagon.

Since all triangles have the same 3 internal angles (x, y, z), they are similar. Since they also have a side of length "a", they must be congruent. Since the side "a" is opposite to angle x in one triangle and opposite to y and z angles in the others, the angles must be the same, x = y = z, which can only happen if they are all 60°. Therefore, the triangles must be equilateral: