Prove that E1 = F1 given that ABCD is a parallelogram.

Given that $ABCD$ is a parallelogram, we know that:

• $AD \parallel BC$ (opposite sides are parallel).
• $AD = BC$ (opposite sides are equal).

• For parallel lines $AD \parallel BC$ and transversal $AC$, $\angle DAE = \angle BCF$ (alternate interior angles).
• For parallel lines $AD \parallel BC$ and transversal $DB$, $\angle ADE = \angle CBF$ (alternate interior angles).

Consider $\triangle AED$ and $\triangle CFB$:

1. $\angle DAE = \angle BCF$ (from Step 2)
2. $AD = BC$ (from Step 1)
3. $\angle ADE = \angle CBF$ (from Step 2) By the Angle-Side-Angle (ASA) congruence criterion, $\triangle AED \cong \triangle CFB$.