1. **State the problem:** Given that BD bisects \(\angle ABC\) and \(\angle ADC\), prove that \(\triangle ABD \cong \triangle CBD\). 2. **Recall the definition of angle bisector:** An angle bisector divides an angle into two equal angles. 3. From the problem, since BD bisects \(\angle ABC\), we have: \[ \angle ABD = \angle DBC \quad \text{(by definition of angle bisector)} \] 4. Similarly, since BD bisects \(\angle ADC\), we have: \[ \angle ADB = \angle CDB \quad \text{(by definition of angle bisector)} \] 5. The segment BD is common to both triangles \(\triangle ABD\) and \(\triangle CBD\), so: \[ BD = BD \quad \text{(Reflexive Property)} \] 6. Now, consider \(\triangle ABD\) and \(\triangle CBD\): - \(\angle ABD = \angle DBC\) (step 3) - \(BD = BD\) (step 5) - \(\angle ADB = \angle CDB\) (step 4) 7. By the Angle-Side-Angle (ASA) congruence postulate, since two angles and the included side are equal, the triangles are congruent: \[ \triangle ABD \cong \triangle CBD \quad \text{(ASA)} \] **Final answer:** \(\triangle ABD \cong \triangle CBD\)