Note that triangle BCA is similar to triangle BHC
Which implies that
HC / BC = CA / BA..... so...
h / a = b / ( x + y) (1)
Now expand ( x + h)^2 + ( y + h)^2 = ( a + b)^2 (2)
x^2 + 2xh + h^2 + y^2 + 2yh + h^2 = a^2 + 2ab + b^2 (3)
And since x^2 + h^2 = a^2 and y^2 + h^2 = b^2
We can subtract these equal parts from (3) and we are left with
2xh + 2yh = 2ab divide through by 2
xh + yh = ab factor out h on the left
h ( x + y) = ab rearrange as
h / a = b / ( x + y) but, by (1)....this is true
So (2) must be true, as well
