1. The problem states that in parallelogram $ABCD$, angle $D$ is a right angle, and we want to prove that the diagonals are congruent. 2. Bao suggests proving this by using the fact that opposite sides in a parallelogram are congruent and by establishing the congruence of a single pair of triangles. 3. The diagonals $AC$ and $BD$ intersect at point $E$. To prove the diagonals are congruent, we look for two triangles that share these diagonals as sides. 4. The triangles to consider are $\triangle ABC$ and $\triangle CDA$ because they share diagonal $AC$ and include sides $AB$ and $DC$, which are opposite and congruent in a parallelogram. 5. Since angle $D$ is a right angle, angle $CDA$ is $90^\circ$. Also, $AB \parallel DC$ and $AD \parallel BC$ imply corresponding angles are congruent. 6. The congruence criterion that fits here is side-angle-side (SAS): two sides and the included angle are congruent. 7. Therefore, Bao is referring to $\triangle ABC$ and $\triangle CDA$ and should use the side-angle-side (SAS) criterion to establish congruence. **Final answer:** Option B: $\triangle ABC$ and $\triangle CDA$ by side-angle-side.