Please provide indepth explination thanks!
Let $\triangle ABC$ be a right triangle, with the point $H$ the foot of the altitude from $C$ to side $\overline{AB}$.
Prove that
Expanding the terms,
\((x^2+2xh+h^2)+(y^2+2yh+y^2)=(a^2+2ab+b^2) \)
Using Pythagorean Theorem,
\(x^2+h^2=a^2\\ y^2+h^2=b^2\)
We could subsitute the values in, and rewrite the equation
\(a^2+2xh+b^2+2yh=a^2+2ab+b^2\\ 2xh+2yh=2ab\\ h(x+y)=ab \)
\([ABC]=\frac12 AB\cdot CH = \frac12 BC \cdot AC\\ \frac12 (x+y)h=\frac12 ab\\ h(x+y)=ab\)
I hope this helped,
Gavin
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