Question: Identify the proof to show that $\triangle ABD \cong \triangle CBD$, where $\angle BDA \cong \angle BDC$ are right angles, $D$ is the midpoint of $AC$, $AB \cong BC$, and $BD$ bisects $\angle B$.

1. **Problem Statement:** We need to prove that $\triangle ABD \cong \triangle CBD$ given: - $\angle BDA \cong \angle BDC$ are right angles. - $D$ is the midpoint of $AC$. - $AB \cong BC$. - $BD$ bisects $\angle B$. 2. **Identify Known Information and What to Prove:** - Since $D$ is midpoint of $AC$, $AD = DC$. - $\angle BDA$ and $\angle BDC$ are right angles, so $\angle BDA = \angle BDC = 90^\circ$. - $AB = BC$ by given. - $BD$ bisects $\angle B$, so $\angle ABD = \angle CBD$. 3. **Choose a Congruence Criterion:** We can use the SAS (Side-Angle-Side) criterion for triangle congruence. 4. **Apply SAS Criterion:** - Side $AB = BC$ (given). - Angle $\angle ABD = \angle CBD$ (since $BD$ bisects $\angle B$). - Side $BD$ is common to both triangles. 5. **Conclusion:** By SAS, $\triangle ABD \cong \triangle CBD$. **Final answer:** $\triangle ABD \cong \triangle CBD$ by SAS congruence criterion.