Final answer:
To prove that the diagonals of a rhombus are perpendicular, we use vector addition and subtraction. By representing the sides of the rhombus as vectors A and B, we show that the diagonals, being A + B and A - B, bisect each other at right angles, thus are perpendicular.
Step-by-step explanation:
To prove that the diagonals of a rhombus are perpendicular, consider each side of the rhombus as a vector. We know that a rhombus has four equal sides, and opposite sides are parallel. When constructing a rhombus using vectors, we can consider the adjacent sides as vectors A and B. The diagonals can then be represented as the resultant vector R, which is A + B, and the difference vector D, which is A - B. Using the properties of parallelograms and parallelogram rule, the diagonals bisect each other at right angles, meaning they are perpendicular. Specifically, the diagonal represented by R being A + B, indicates that it splits the rhombus into two congruent triangles, and hence it bisects angle A evenly. The same applies for diagonal D = A - B. Since vectors A and B are perpendicular, their sum and difference will also intersect at a 90° angle, proving that the diagonals are indeed perpendicular.