Final answer:
Diagonals in a rectangle divide it into four congruent triangles by the SSS congruence postulate. The intersection point of the diagonals serves as the midpoint for both, making the segments they create equal in length.
Step-by-step explanation:
To prove that both diagonals divide the rectangle into two pairs of congruent triangles, consider a rectangle ABCD with diagonals AC and BD intersecting at point M. A rectangle has congruent opposite sides and right angles at each vertex. The diagonals of a rectangle are congruent as well. By drawing diagonals AC and BD, we have created four triangles: ΔAMB, ΔCMD, ΔAMC, and ΔBMD.
In triangles ΔAMB and ΔCMD, side AB is congruent to side CD (opposite sides of a rectangle are congruent), side AM is congruent to side CM (halves of diagonal AC), and side BM is congruent to side DM (halves of diagonal BD). By the SSS postulate, the triangles are congruent. Similarly, triangles ΔAMC and ΔBMD can be proven congruent using the same reasoning.
To prove that M is the midpoint of both diagonals, note that in a rectangle, the diagonals bisect each other. So if M is where diagonals AC and BD intersect, AM is congruent to CM, and BM is congruent to DM. This definitionally means that M is the midpoint of both diagonals.
Thus, both diagonals divide the rectangle into two pairs of congruent triangles (ΔAMB ≅ ΔCMD and ΔAMC ≅ ΔBMD), and point M is the midpoint of both diagonals AC and BD (i.e., M divides AC and BD into equal parts).