The way this problem and the recent circle in the triangle in the semicircle problem, are posed signals that the answer does not depend upon the shape of the original (or only) triangle. Thus we are free to take the simplest case and assume the answer applies general (which IS kind of cool).

In this case make the original triangle equilateral of length one, so that the area of each inner square is 1.

Now bisect the outer triangles into two right triangle (each). Its hyponeuse and short leg are 1 and 1/2 respectively so the long leg is Sqr(3)/2, Yielding an area for each outer square as 3. Which is the answer!

So is this really true for all original triangles?? Try another extreme case; a "triangle" with two sides 1 long and the third 2 long. The areas of the inner squares add up to 6. The area of the top outer square is zero. The two side square are 3 by 3, totaling 18 BINGO!!