In Fig. AB and FE , If BC=DE and AB=EF, then is congruent to: - Vidyadip

MCQ

In Fig. $\text{AB}\perp\text{BE}$ and $\text{FE}\perp\text{BE},$ If $\text{BC}=\text{DE}$ and $\text{AB}=\text{EF},$ then $\triangle\text{ABD}$ is congruent to:

A
$\triangle\text{EFC}$
B
$\triangle\text{ECF}$
C
$\triangle\text{CEF}$
D
$\triangle\text{FEC}$

✓

Answer

$\triangle\text{FEC}$
Solution:

AB = EF
BC = DE
BC + CD = DE + CD (adding CD both sides)
BC + CD = BD, DE + CD = CE
So BD = CE
Now Consider $\triangle\text{ABD},$ & $\triangle\text{FEC}$
$\text{AB}=\text{FE}$
$\text{BD}=\text{EC}$
$\angle\text{ABD}=\angle\text{FEC}=90^\circ$
So $\triangle\text{ABD}\cong\triangle\text{FEC}$ by SAS creterion.
Hence, correct option is (d).

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