# Math Help - Proof

1. ## Proof

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

2. Here is a simple solution:

$Join A and D.$

$AE=BE=CE=DE$

Diagnals of quadrilateral $ABCD$ bisect each other.
Diagnals of a parallelogram bisect each other.
So $ABCD$ is a parallelogram.
So AB=CD.
In $\triangle{ABC}$and $\triangle{BCD}$,
BC is common side;
AC=BD
and AB=CD.
SO $\triangle{ABC} \cong{DCB}$

3. ## Its not a parallelogram

I cant say anything aboiut a pallogram because we are missing a whole side of the parallelogram, ok I see I didnt see you say join AD