Assuming that vertices ABCD are in clockwise order around the rectangle:
First, I note that triangles ACB and CAB have the same vertices, and hence are the SAME TRIANGLE, and hence are congruent.
But we can also show congruence of triangle ACB and ACD.
Because it's a triangle, all corners of the rectangle are right angles.
Angle ABC and ADC are opposite corners of the rectangle, and are both right angles (and therefore equal).
Note that line segments AB and DC are parallel, due to the fact that it's a rectangle.
Since AB and DC are parallel, and diagonal AC intersects both of them, angle CAB is congruent to angle ACD.
Since, triangles ACB and ACD have 2 congruent angles, then the remaining angles in each triangle (CAD and ACB) are also congruent. Triangles ACB and ACD are therefore similar.
Since it's a rectangle, sides AB and DC are congruent.
Since we have 2 similar triangles that have at least one side that is congruent, then the 2 triangles ACB and ACD are congruent.