Draw the 5-12-13
Draw the three inside lines. Two should be easy. The smaller hypotenuse may be approximate. Don't worry about that right now.
From each of the inner-hypotenuse endpoints (where it intersects the inner-legs, draw line segments parallel to the legs.
This should result in a nice collection of other similar triangles, of various sizes.
Now what?
Just to emphasize that unique answers don't care how you get them, I used an analytic geometry solution.
Line 1: x = 0
Line 2: y = 0
Line 3: x/5 + y/12 = 1 ==> y = 12(1-x/5) = 12 - (5/12)x
That's the outer triangle.
Now the inner triangle...
Line 4: x = 1
Line 5: y = 1
Line 6: This is the only tricky part. I did it this way.
Line 7: Through the Origin and perpendicular to Line 3 -- y = (12/5)x Intersects Line 3 at (720/169,1728/169). This intersection is 144/13 from the Origin - just using the distance formula.
Line 6: Is parallel to Line 3 and intersects Line 7 at a distance of 144/13 - 1 = (144-13)/13 = 131/13 from the Origin.
You can finish, if you like. It's almost done. This may seem massive and confusing or perhaps far more complicated than it has to be. That's okay. There is OFTEN more than one way to go about it. Some ways are really awesome and slick. Other ways are cumbersome and entirely laborious. The answer is still unique! Sometimes, just keeping tack through a swampy mess of algebra can be quite rewarding.
