Proving diagonals of rhombus are perpendicular using coordinate geometry: why prove c^2 = a^2 - b^2?

Question

How do you prove the diagonals of a rhombus are perpendicular?

On this problem I have determined the points should be (0,0),
(a,0) (b,c) and (b+a,c). I have also found the slopes of the diagonals to be c/b+d and c/(b-d). But I don't what to do after this. I think you would prove that c^2 = a^2 - b^2 using cross multiplication on the slopes, but why would you want to prove this? I thought that opposite reciprocal slopes signify perpendicular lines; does this signify that too?

I don't see why getting c^2 = a^2 - b^2 help me on proving the diagonals are perpendicular

Answer 1

For simplicity let's label the four vertices.

O(0,0)
K(a,0)
L(b,c)
M(b+a,c)

The one thing you forgot about is that all the sides of a rhombus are of equal length. So drop a perpendicular from L to the point P(b,0) on side OK. Now you have a right triangle with:

legs
OP = P(b,0) - O(0,0) = b
LP = L(b,c) - P(b,0) = c

hypotenuse
OL = L(b,c) - O(0,0) = √(b² + c²) = a
since all sides of a rhombus are of equal length

b² + c² = a²
c² = a² - b²

And you can take it from there.

Answer 2

The Pythagorean equality is true if and only if the triangle is a right triangle.

Is that the hint you're looking for?