Two congruent rhombi share the point of intersection of their diagonals. Shorter diagonals are perpendicular to each other. Prove that the line joining the point of intersection of the diagonals and the midpoint of the side of one rhombus is perpendicular to a side of the other rhombus

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Explanation:

Without loss of generality, we can let the intersection of the diagonals be the origin of our coordinate system, and the diagonals align with the x- and y-axes. We can define the half-diagonals of rhombus ABCD as 'a' and 'b', representing the distances from the origin to vertices A and B, respectively.

With A on the -x axis and vertices named clockwise, the slope of side AB will be b/a. When the rhombus is rotated 90° CCW, the slope of the side B'C' becomes a/b.

The midpoint of side AB is located at (-a/2, b/2), so the slope of the line from there through the origin is (b/2)/(-a/2) = -b/a.

So far, we have the slope to the midpoint of one side of the original rhombus is -b/a, and the slope of the side of the rotated rhombus is a/b. The product of these slopes is (-b/a)(a/b) = -1, so the two segments are perpendicular. So, we have shown a line through the center of a rhombus and the midpoint of one side is perpendicular to the side of the rhombus after it is rotated 90° about its center.

cosine

step-by-step explanation: