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Question
In the given figure, AB ⊥ BE and FE ⊥ BE. If BC = DE and AB = EF, then ΔABD is congruent to
Options
ΔEFC
ΔECF
ΔCEF
ΔFEC
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Solution
ΔFEC
Explanation:
It is given that
In the figure, AB ⊥ BE, FE ⊥ BE
Now, BC = DE.
Adding DC to both the sides, we get,
⇒ BC + DC = DE + DC
⇒ BD = CE
In ΔABD and ΔCEF
BD = CE (Proved)
AB = FE (Given)
∠ABD = ∠FEC (Each 90°)
∴ ΔABD ≅ ΔFEC by SAS congruence rule.
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