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