Given: Line AC is parallel to DF, Line BE is perpendicular to DF, and angle AEB is congruent to angle CEB, prove angle BAE is congruent to angle BCE. Will give Brainliest if explained thoroughly.

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Explanation:

There are several ways you can go at this. Here are a couple. All proofs will start with the given relations being repeated as part of the proof. Here are the next steps.

∠AED ≅ ∠BAE . . . . alternate interior angles are congruent

∠AED +∠AEB = 90° . . . . angle sum theorem

∠BAE +∠AEB = 90° . . . . substitution property of equality

∠CEF ≅ ∠BCE . . . . alternate interior angles are congruent

∠CEF +∠CEB = 90° . . . . angle sum theorem

∠BCE +∠CEB = 90° . . . . substitution property of equality

∠BAE +∠AEB = ∠BCE +∠CEB . . . . substitution property of equality

∠BAE +∠AEB = ∠BCE +∠AEB . . . . substitution property of equality

∠BAE = ∠BCE . . . . addition property of equality

∠ABE = ∠CBE = 90° . . . . BE ⊥ AC

BE ≅ BE . . . . reflexive property of congruence

ΔBEA ≅ ΔBEC . . . . ASA congruence theorem

∠BAE ≅ ∠BCE . . . . CPCTC