Geometry - Similar Triangles again

For spring break here’s a geometry walk through I wrote up a while back but never got around to publishing.

Another walk through of a @five_triangles problem.  Given the following diagram find

the length of AB.

Total Time: 3 days (on bus, before bed)

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.

1. This also created a 3rd smaller similar triangle.

2. Now I just needed one of the side lengths in the new triangle.

3. 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.

4. 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.