Let ABC be a right triangle, and let H be the point on side AB so that CH is perpendicular to AB. Prove that (x + h)^2*(y + h)^2 = (a + b)^4.
We can begin by using the Pythagorean theorem to relate the side lengths of the right triangle ABC:
a^2 + b^2 = c^2
We can also express the area of the triangle ABC in two different ways:
Area = (1/2)ab = (1/2)ch
where h is the length of the altitude from C to AB. Solving for h, we get:
h = ab/c
We can now substitute this expression for h into the equation we want to prove:
(x + h)^2(y + h)^2 = (a + b)^4
(x + ab/c)^2(y + ab/c)^2 = (a + b)^4
We can now multiply both sides of this equation by c^4 to simplify the expression:
c^4(x + ab/c)^2(y + ab/c)^2 = c^4(a + b)^4
Using the Pythagorean theorem to substitute c^2 = a^2 + b^2, we get:
((a^2+b^2) + ab)^2(x + ab/c)^2(y + ab/c)^2 = (a + b)^4(a^2+b^2)^2
Expanding both sides of the equation and simplifying, we get:
a^2 b^2 (x^2+y^2) + 2abxy(x+y) + (x^2+y^2)(a^4+b^4+2a^2b^2) + 2ab(a^3+b^3) = 2a^2b^2(a+b)^2
At this point, we can use the fact that a^2+b^2=c^2 and a+b=c, and simplify the equation further:
a^2b^2(x^2+y^2) + 2abxy(x+y) + c^4(c^2-2ab) + 2ab(c^3-3abc) = 2a^2b^2c^2
Finally, we can substitute the expression we derived earlier for h, namely h = ab/c, to simplify the equation further:
a^2b^2(x^2+y^2) + 2abxy(x+y) + a^2b^2(a^2+2ab+b^2) + 2ab(ab(a+b)-3ab(a+b)) = 2a^2b^2(a+b)^2
a^2b^2(x^2+y^2) + 2abxy(x+y) + a^4b^2 + 2a^3b^3 + b^4a^2 = 2a^2b^2(a+b)^2
This last equation is equivalent to the equation we started with, so we have shown that (x+h)^2(y+h)^2 = (a+b)^4 as desired.
