Given
[tex]\bar{AB}\mleft\Vert\bar{DC}\mright?[/tex]
and E is the midpoint of AC.
Therefore,
[tex]\bar{AE}\cong\bar{EC}\text{ (Reason: E is the midpoint of AC)}[/tex]
Since [tex]\angle AEB\cong\angle DEC\text{ (Reason: Vertical angles are congruent)}[/tex][tex]\angle ABE\cong\angle EDC\text{ (Reason: Parallel lines cut by a transversal form congruent alternate interior angles)}[/tex]
Therefore,
[tex]\triangle\text{AEB }\cong\triangle\text{DEC (Reason: It supports the AAS congruency rule)}[/tex][tex]\angle AEB=90\degree\text{ (Reason: Intersecting transversals are perpendicular to each other)}[/tex]
Therefore,
[tex]\triangle\text{AEB and }\Delta DEC\text{ are right angled triangles}[/tex]
