Question
In the following figure, $\text{AB}=\text{EF},\text{BC}=\text{DE}$ and $\angle B=\angle E=90^{\circ}$.
Prove that $\text{AD}=\text{FC}$
Prove that $\text{AD}=\text{FC}$
✓
Answer
Given that, $\mathrm{BC}=\mathrm{DE}$
$\Rightarrow \text{BC}+\text{CD}=\text{DE}+\text{CD} \ldots .($ Adding $\text{CD}$ on both sides $)$
$\Rightarrow \mathrm{BD}=\mathrm{CE} ....(i)$
Now, in $\triangle \mathrm{ABD}$ and $\triangle \mathrm{FEC}$,
$\mathrm{AB}=\mathrm{EF} ....($given$)$
$\angle \mathrm{ABD}=\angle \mathrm{FEC}$
$\ldots\left(\right.$ Each $\left.90^{\circ}\right)$
$\text{BD}=\text{CE} ....[$ From $(i) ]$
$\Rightarrow \triangle \mathrm{ABD} \cong \triangle \mathrm{FEC} ...($by $\text{SAS}$ congruence criterion$)$
$\Rightarrow \text{AD}=\text{FC} ...($c.p.c.t.$)$
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