Another walk through of a @five_triangles problem. Given the following diagram find
the length of AB.
In retrospect I circled around the solution and it seems much more obvious than it was while working on the problem.
1. First thought was that we'd definitely need a auxiliary line but my first idea was to extend the triangle containing AB.
2. Secondly I noticed the 30-60-90 triangles with the 2cm-4cm sides.
3. I calculated the other side lengths in the right triangles using the Pythagorean Theorem.
4. I wondered if AB was congruent with the 4cm side and if I could prove it.
5. Then I noticed an isosceles 3 - 3 triangle but that didn't add any other angles.
6. I did some angle chasing to see if I could show AB was 4 and in the process realized I had another
big similar triangle. If I extended the other line.
7. This also created a 3rd smaller similar triangle.
8. Now I just needed one of the side lengths in the new triangle.
9. Because I knew a bunch of the original base lengths I started looking for ways to find the rest which would give a side of the similar triangles. But this became a dead end.
9. Then I realized the hypotenuse was 4 + the hypotenuse of the littler triangle and the littler triangle
had enough side lengths to figure the rest out.