Question

given \\( \overline{adb} \\), \\( \overline{bec} \\), \\( \overline{bd} \cong \overline{be} \\), \\( \overline{da} \cong \overline{ec} \\)
prove \\( \angle 1 \cong \angle 2 \\)

Explanation:

Step1: Prove \(\triangle ABD\cong\triangle CBE\)

Since \(\overline{ADB}\), \(\overline{BEC}\), \(\overline{BD}\cong\overline{BE}\), \(\overline{DA}\cong\overline{EC}\), by the Side - Side - Side (SSS) congruence criterion (\(AB = CB\) (because \(AD + DB=EC + EB\) and \(AD = EC\), \(DB = EB\)), \(BD = BE\), \(DA=EC\)), \(\triangle ABD\cong\triangle CBE\).

Step2: Use congruent triangles property

Because \(\triangle ABD\cong\triangle CBE\), then \(\angle BAD=\angle BCE\).

Step3: Prove \(\triangle AFD\cong\triangle CFE\) (where \(F\) is the intersection of \(AC\) and \(DE\))

\(\angle AFD=\angle CFE\) (vertical angles), \(\angle BAD=\angle BCE\), \(DA = EC\). By the Angle - Angle - Side (AAS) congruence criterion, \(\triangle AFD\cong\triangle CFE\).

Step4: Get the result

Since \(\triangle AFD\cong\triangle CFE\), then \(\angle 1=\angle 2\) (corresponding angles of congruent triangles).

Answer:

\(\angle 1\cong\angle 2\) is proved as above.