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# Help with proof

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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:

$$(x + h)^2 + (y + h)^2 = (a + b)^2$$

$$x^2+2xh+h^2+y^2+2yh+h^2=a^2+2ab+b^2$$

$$x^2+2xh+h^2+y^2+2yh+h^2=x^2+h^2+2ab+y^2+h^2$$

Nov 1, 2018

#1
+102948
+3

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      ....!!!!

Nov 2, 2018
edited by CPhill  Nov 2, 2018
#2
+37
+2

Thank you so much. I guess i couldnt see the forest trough the trees

bambam89  Nov 2, 2018
#3
+102948
+2

Well...to be honest....the first time I encountered this one it kind of "threw" me, too.  But, sometimes, just one little "trick"  is the key to solving a problem.....in this case..the "similar triangles" thing...

CPhill  Nov 2, 2018