## Elementary Geometry for College Students (6th Edition)

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