+37 Let $\triangle ABC$ be a right triangle, with the point $H$ the foot of the altitude from $C$ to side $\overline{AB}$.
[asy]
// Coordinates for 3-4-5 triangle
pair A = (5, 0);
pair B = (0, 0);
pair C = (1.8, 2.4);
pair H = foot(C, A, B);
draw(A--B--C--cycle);
draw(rightanglemark(B, C, A));
draw(C--H);
draw(rightanglemark(C,H,A));
label("$A$", A, E);
label("$B$", B, W);
label("$C$", C, N);
label("$H$", H, S);
label("$x$", midpoint(B--H), S);
label("$y$", midpoint(A--H), S);
label("$h$", midpoint(C--H), E);
label("$a$", midpoint(B--C), NW);
label("$b$", midpoint(A--C), NE);
[/asy]
Prove that
\[(x + h)^2 + (y + h)^2 = (a + b)^2.\]
I got to \([\frac{2xh+2yh=2ab}{2}]=[xh+yh=ab]\) but idk wat to do after that. Here is my work:
\((x + h)^2 + (y + h)^2 = (a + b)^2\)
\(x^2+2xh+h^2+y^2+2yh+h^2=a^2+2ab+b^2\)
\(x^2+2xh+h^2+y^2+2yh+h^2=x^2+h^2+2ab+y^2+h^2\)
+128475 I think this might be similar to what you need :
Note that triangles BHC and BCA are similar...so...
HC / BC = CA / BA so
h / a = b / ( x + y) (1)
Working with your expansion...note that
x^2 + h^2 = a^2 and
y^2 + h^2 = b^2
So...subtracting out the equal parts we are left with
2xh + 2yh = 2ab (3) divide through by 2
xh + yh = ab factor the left side
h ( x + y) = ab which we can rearrange as
h / a = b / ( x+ y) (2)
And since this is equal to (1) and it's just a rearrangement of (3) adding all the other equal parts of the expansion shows that
(x^2 + h^2) + (y^2 + h^2) + ( 2xh + 2yh) = a^2 + b^2 + 2ab
( x^2 + 2xh + h^2) + ( y^2 + 2yh + h^2) = ( a + b)^2
( x + h)^2 + ( y + h)^2 = (a + b)^2 ....!!!!
