geometry

\(\overline{AB} \cong \overline{AD}\\ \overline{CB} \cong \overline{CD}\\ \overline{AC} \cong \overline{AC}\\ \)

Therefore by SSS postulate, \(\triangle ABC \cong \triangle ADC\).

By congruent triangles, we know that \(m\angle BAE = m\angle DAE \) --- (1)

Also, \(\overline{AB} \cong \overline{AD}\) --- (2)

and \(\overline{AE} \cong \overline{AE}\\\) --- (3)

 

By (1), (2), (3), and SAS postulate, \(\triangle BAE \cong \triangle DAE\) --- (4)

Then, by (4), \(\angle BEA = \angle DEA\) --- (5)

Also, by sum of angles on a straight line, \(\angle BEA + \angle DEA = 180^\circ\) --- (6)

Solving (5) and (6) gives \(\angle BEA = \angle DEA = 90^\circ\)

 

Therefore, \(\triangle ABE \) is a right triangle with \(\angle BEA = 90^\circ\).