#### Answer

(a) By AAS, $\triangle ABD \cong \triangle ACD$
(b) By SSS, $\triangle ABD \cong \triangle ACD$
(c) By SAS, $\triangle ABD \cong \triangle ACD$

#### Work Step by Step

(a) Angle: $\angle ADC \cong \angle ADB$, since both angles are $90^{\circ}$ by definition of altitude
Angle: $\angle ACD \cong \angle ABD$, since $\overline{AB} \cong \overline{AC}$
Side: $\overline{AB} \cong \overline{AC}$, since it is given in the question
Therefore, by AAS, $\triangle ABD \cong \triangle ACD$
(b) Side: $\overline{AB} \cong \overline{AC}$, since it is given
Side: $\overline{AD} \cong \overline{AD}$, by identity since it is the same line
Side: $\overline{BD} \cong \overline{CD}$, by definition of median
Therefore, by SSS, $\triangle ABD \cong \triangle ACD$
(c) Side: $\overline{AB} \cong \overline{AC}$, since it is given in the question
Angle: $\angle BAD \cong \angle CAD$, by definition of bisector
Side: $\overline{AD} \cong \overline{AD}$, by identity since it is the same line
Therefore, by SAS, $\triangle ABD \cong \triangle ACD$