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Question

In the following figure AB = EF , BC = DE and $$ \angle B = \angle E = 90^{o}$$
Prove that AD = FC

Solution
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Given that, $$BC = DE$$
$$ \Rightarrow BC + CD = DE + CD$$ (Adding $$CD$$ on both sides)
$$\Rightarrow BD = CE $$ ...(1)
Now, In$$ \Delta ABD$$ and $$\Delta EFC $$
$$ AB = EF $$ ...[Given]
$$\angle ABD = \angle FEC $$ ...[Both $$=90^0$$, perpendiculars]
$$ BD = CE $$ ...[From (1)]
$$\Rightarrow ABD \cong \Delta \Delta EFC $$ ...[SAS congruence criterion]
$$ \Rightarrow AD = FC $$ ...[cpct]

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