Let \(\triangle ABC\) be a right triangle, and let \(H\) be the point on side \(\overline{AB}\) so that \(\overline{CH} \perp \overline{AB}.\)
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Prove that \((x + h)^2 + (y + h)^2 = (a + b)^2.\)
Thank you!
Find the area of the triangle in two different ways:
First way: Area = ½·base·height ---> Area = ½·(x + y)·h
Second way: Area = ½·a·b·cos(C) ---> Area = ½·a·b [cos(90°) = 1]
Therefore: ½·(x + y)·h = ½·a·b
Multiply both sides by 4: 2·(x + y)·h = 2·a·b
Add (x2 + y2) to both sides: (x2 + y2) + 2·(x + y)·h = (x2 + y2) + 2·a·b
Simplify: (x2 + y2) + 2xh + 2yh = (x2 + y2) + 2·a·b
Rewrite: (x2 + 2xh) + (y2 + 2yh) = (x2 + y2) + 2·a·b
Add 2h2 to both sides: (x2 + 2xh + h2) + (y2 + 2yh + h2) = (x2 + y2) + 2·a·b + 2h2
Factor: (x + h)2 + (y + h)2 = (x2 + y2) + 2·a·b + 2h2
Complete the square: (x + h)2 + (y + h)2 = (x2 + 2xy + y2) + 2·a·b + 2h2 - 2xy
Rewrite: (x + h)2 + (y + h)2 = (x + y)2 + 2·a·b + 2h2 - 2xy
Since (x + y)2 = c2 = a2 + b2 (x + h)2 + (y + h)2 = (a2 + b2) + 2·a·b + 2h2 - 2xy
Rewrite: (x + h)2 + (y + h)2 = (a2 + 2ab + b2) + 2h2 - 2xy
Rewrite: (x + h)2 + (y + h)2 = (a + b) 2+ 2h2 - 2xy
Going back to the triangle: h is the mean proportion between x and y: x/h = h/y --> xy = h2
Therefore 2xy = 2h2 ---> 2h2 - 2xy = 0
So: (x + h)2 + (y + h)2 = (a + b) 2+ 0
And: (x + h)2 + (y + h)2 = (a + b) 2