In the figure, $AB\perp BE$ and $FE\perp BE$. If $BC=DE$ and $AB=EF$, then $\Delta ABD$ is congruent to

The correct option is C. $\Delta FEC$

In the figure, $AB\perp BE,FE\perp BE$

Now, $BC=DE$.

Adding $DC$ to both the sides, we get,

$\Rightarrow BC+DC=DE+DC$

$\Rightarrow BD=CE$

In $\Delta ABD$ and $\Delta CEF.$

$BD=CE$ (Proved)

$AB=FE$ (Given)

$\angle ABD=\angle FEC$ (Each ${90}^{\circ }$)

$\therefore \Delta ABD\cong \Delta FEC$ by SAS congruence rule.