In the given figure, AB ⊥ BE, EF ⊥ BE, AB = EF and BC = DE, then :

1. △ ABD ≅ △ EFC

2. △ ABD ≅ △ FEC

3. △ ABD ≅ △ ECF

4. △ ABD ≅ △ CEF

Since, AB ⊥ BE and EF ⊥ BE,

∴ ∠ABE = 90° and ∠FEB = 90°.

From figure,

⇒ ∠ABD = ∠ABE = 90°

⇒ ∠FEC = ∠FEB = 90°

Given,

⇒ BC = DE

⇒ BC + CD = CD + DE

⇒ BD = CE.

In △ ABD and △ FEC,

⇒ AB = FE (Given)

⇒ BD = CE (Proved above)

⇒ ∠ABD = ∠FEC (Both equal to 90°)

∴ △ ABD ≅ △ FEC (By S.A.S. axiom)

Hence, Option 2 is the correct option.