+0  
 
0
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3
avatar+28 

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:

 

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

 

\(x^2+2xh+h^2+y^2+2yh+h^2=a^2+2ab+b^2\)

 

\(x^2+2xh+h^2+y^2+2yh+h^2=x^2+h^2+2ab+y^2+h^2\)

bambam89  Nov 1, 2018
 #1
avatar+90968 
+2

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      ....!!!!

 

 

cool cool cool

CPhill  Nov 2, 2018
edited by CPhill  Nov 2, 2018
 #2
avatar+28 
+1

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

bambam89  Nov 2, 2018
 #3
avatar+90968 
+1

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...

 

 

cool cool cool

CPhill  Nov 2, 2018

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