Question

given: \\( \overline { a b } \cong \overline { b c } \\) and \\( e \\) is the midpoint of \\( \overline { a c } \\).
prove: \\( \angle b a d \cong \angle b c d \\).

Explanation:

Step1: Consider triangles \( \triangle ABE \) and \( \triangle CBE \)

Since \( \overline{AB}\cong\overline{BC}\), \( \overline{AE}\cong\overline{CE}\) (from step 2), and \( \overline{BE}\cong\overline{BE}\) (common side). By SSS (Side - Side - Side) congruence criterion, \( \triangle ABE\cong\triangle CBE\). So, \( \angle BAE\cong\angle BCE\).

Step2: Consider triangles \( \triangle ADE \) and \( \triangle CDE \)

Since \( \overline{AE}\cong\overline{CE}\) (from step 2), \( \overline{DE}\cong\overline{DE}\) (common side), and \( \angle AED=\angle CED = 90^{\circ}\) (because \( \triangle ABE\cong\triangle CBE\), so \( BE\perp AC\) and \( \angle AED\) and \( \angle CED\) are right angles). By SAS (Side - Angle - Side) congruence criterion, \( \triangle ADE\cong\triangle CDE\). So, \( \angle DAE\cong\angle DCE\).

Step3: Use angle addition

\(\angle BAD=\angle BAE+\angle DAE\) and \(\angle BCD=\angle BCE+\angle DCE\). Since \( \angle BAE\cong\angle BCE\) and \( \angle DAE\cong\angle DCE\), by the angle addition property and substitution, \( \angle BAD\cong\angle BCD\).

Answer:

\(\angle BAD\cong\angle BCD\) is proved as above.