Hello, Medusa!

Can anyone explain how the formula: bh/(b+h) works in this situation?

b = base, h = height. Code:

- - Ao
: : | *
: h-x | *
: : | *
: - Do-------oE
h : | x | *
: : | | *
: x | x| *
: : | | *
: : | | *
- - o-------o-------------o
B - x - F - - b-x - - C
: - - - - b - - - - :

We have right triangle *ABC* with *AB = h, BC = b.*

We have square *BDEF* with sides *x.*

Since ∆ABC ~ ∆ADE, we have:

. . .h. . . .h - x

. . --- .= .----- . . → . . hx .= .b(h - x) . . → . . hx .= .bh - bx

. . .b. . . . . x

. . bx + hx .= .bh . . → . . (b + h)x .= .bh

. . . . . . . . . . . . . . .bh

Therefore: . x .= .-------

. . . . . . . . . . . . . .b + h