Answer:
See proof below
Step-by-step explanation:
Given [tex]\triangle ADE \cong \triangle CBD[/tex]
Then, [tex]\angle AED \cong \angle CFB[/tex] because corresponding parts of congruent triangles are congruent.
Since [tex]\angle AED \text{ and } \angle DEB \text{ form a linear pair}[/tex], by the linear pair postulate, [tex]\angle AED \text{ and } \angle DEB \text{ are supplementary}[/tex].
Similarly, [tex]\angle CFB \text{ and } \angle BFD \text{ form a linear pair}[/tex], so by the linear pair postulate, [tex]\angle CFB \text{ and } \angle BFD \text{ are supplementary}[/tex].
By the Congruent supplements theorem, since [tex]\angle AED \text{ and } \angle DEB \text{ are supplementary}[/tex], [tex]\angle CFB \text{ and } \angle BFD \text{ are supplementary}[/tex], and [tex]\angle AED \cong \angle CFB[/tex] , then [tex]\angle DEB \cong \angle BFD[/tex]. (note, this is one pair of opposite angles inside quadrilateral DFBE)
Recalling that [tex]\overline {ED} \text{ bisects } \angle ADC, \text{ and } \overline {FB} \text{ bisects } \angle ABC[/tex], then, [tex]\angle ADE \cong \angle CDE[/tex], and [tex]\angle CBF \cong \angle FBA[/tex] by definition of angle bisector.
Note that [tex]\angle ADE \cong \angle CBF[/tex] because corresponding parts of congruent triangles are congruent.
Also, note that [tex]\angle EDF \cong \angle EDC[/tex] and [tex]\angle FBA \cong \angle FBE[/tex] because B, E, A are colinear, and D, F, C are colinear.
So, by the transitive property of angle congruence, [tex]\angle EDF \cong \angle FBE[/tex] (This is the other pair of opposite angles inside quadrilateral DFBE)
So, since both pairs of opposite angles are congruent, quadrilateral DFBE is a parallelogram, by a theorem about quadrilateral properties (your book may or may not have a name for it. It may just have a number).
