I'm going to warehouse these problems from @ five_triangles here. I really like how they both show constructions for a 3-4-5 Pythagorean Triple. My plan is to keep looking for other variants on this theme. This is almost enough material to make a followup day after the Pythagorean Theorem although it still needs a few slightly easier problems to go with it.

#### Discussion

Examine triangle DCO.- DC is R + r
- DO is 2R - r
- OC is R

Apply the Pythagorean Theorem: $$ (R+r)^2 = (2R - r)^2 + R^2$$

After simplifying this reduces to $$3Rr = 2R^2$$ and then assuming R is positive $$ r = \frac{2}{3} R$$ Or put another way: r is 1/3 of the side and R is 1/2 of the side. Plug those values back into the original triangle and voila you find the 3-4-5.

After simplifying this reduces to $$3Rr = 2R^2$$ and then assuming R is positive $$ r = \frac{2}{3} R$$ Or put another way: r is 1/3 of the side and R is 1/2 of the side. Plug those values back into the original triangle and voila you find the 3-4-5.

#### Discussion

This construction is a bit trickier.

- Note the congruent segments in the two big triangles
- Angle chase to find the 2 sets of similar triangles and the right angles.
- From the first set AFC and AFE and the given hypotenuse lengths You can immediately see the areas are in a ratio of 4:1 as is BF:EF since they share an altitude AF.
- Let H be the intersection of GC and the line from B. This is a right angle from the initial angle chase. This second set EFG and EBH are also similar. For EBH you know have a right triangle with a hypotenuse of 5x and one side of 4x from step 3 where x is the length of EF. The Pythagorean theorem immediately says EH must be 3x.
- Its easy to do the 3:1 scaling if you want to find the area of EFG from there.

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