Triangles within a Parallelogram ABCD is a parallelogram.
E is the point where the diagonals AC and BD meet.
Prove that triangle ABE is congruent to triangle CDE.

AE and EC are on the same line, so have the same gradients. Same goes for BE and ED. Because it is a parallelogram, AB=CD, as it has to be so that the shape holds. Therefore, you have proved this by the Angle-Side-Angle rule, where two triangles with two identical angles and a side are congruent, even if they are reflected.
If you know the gradient, and set the known side to 0 degrees, the gradient shows the vector, which is also an angle.
