Given AE congruent BE congruent CE congruent DE
Prove Triangle ABC congruent triangle DCB
Am I able to say that angle a is congruent to angle due the sides that they are across from all being congruent
Here is a simple solution:
$\displaystyle Join A and D. $
$\displaystyle AE=BE=CE=DE$
Diagnals of quadrilateral $\displaystyle ABCD$ bisect each other.
Diagnals of a parallelogram bisect each other.
So$\displaystyle ABCD$ is a parallelogram.
So AB=CD.
In $\displaystyle \triangle{ABC} $and $\displaystyle \triangle{BCD}$,
BC is common side;
AC=BD
and AB=CD.
SO $\displaystyle \triangle{ABC} \cong{DCB}$