+37 Let △ABC be a right triangle, with the point H the foot of the altitude from C to side ¯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 [2xh+2yh=2ab2]=[xh+yh=ab] but idk wat to do after that. Here is my work:
(x+h)2+(y+h)2=(a+b)2
x2+2xh+h2+y2+2yh+h2=a2+2ab+b2
x2+2xh+h2+y2+2yh+h2=x2+h2+2ab+y2+h2
+130477 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 ....!!!!
