Suppose ABCD is a rectangle. Using the RHS theorem proves that triangles ABC and ADC are congruent.

Hint:
We will prove congruence using the Right angle – Hypotenuse – Side theorem. We will identify a right angle, a hypotenuse of the same length and a side in both the triangles. If we are able to identify these 3 things, the 3 criteria of the RHS theorem will be satisfied and the 2 triangles will be proved to be congruent.

Complete step by step solution:
Two geometrical figures are said to be congruent if they both have the same shape and same size.
The RHS congruence rule states that if 2 right-angled triangles have a hypotenuse of the same length and one side (other than the hypotenuse) of the same length, then the 2 triangles are congruent.

In \[\Delta {\rm{ABC}}\] and \[\Delta {\rm{ADC}}\] :-
\[\angle {\rm{ABC}} = \angle {\rm{ADC}}\] (Both the angles are right angles because all angles of a rectangle are right angles.)
\[{\rm{AC}} = {\rm{AC}}\] (The hypotenuse of both the triangles is common and thus equal. It is the diagonal of the rectangle.)
\[{\rm{AB}} = {\rm{DC}}\] (Sides AB and DC of the triangles are equal because they form opposite sides of the rectangle ABCD and opposite sides of a rectangle are always equal.)
As \[\Delta {\rm{ABC}}\] and \[\Delta {\rm{ADC}}\]satisfy all the three criteria of the RHS theorem, they are congruent triangles.

\[\therefore \] \[\Delta {\rm{ABC}} \cong \Delta {\rm{ADC}}\]

Note:
When we are proving congruence using the RHS theorem, we can choose any side of the right-angled triangle. It can be the base of the triangle or the perpendicular of the triangle. In this question, we can also prove congruence using sides BC and AD instead of AB and DC in the third step.