Question
In $\text{BA}\perp\text{AC},\text{DE}\perp\text{DF}$ such that $BA = DE$ and $BF = EC$. Show that $\triangle\text{ABC}\cong\triangle\text{DEF.}$
✓
Answer
Given, $\text{BA}\perp\text{AC},\text{DE}\perp\text{DF}$ such that $BA = DE and BF = EC$.$\triangle\text{ABC}\cong\triangle\text{DEF}$
$\text{BF}=\text{EC}$
On adding CF both sides, we have
$\text{BF}+\text{CF}=\text{EC}+\text{CF}$
$\text{BC}=\text{EF}$
In $\triangle\text{ABC}$ and $\triangle\text{DEF},$
$\angle\text{A}=\angle\text{D}=90^{\circ}$
$\text{BA}=\text{DE}$
$\triangle\text{ABC}\cong\triangle\text{DEF}$
$\text{BF}=\text{EC}$
On adding CF both sides, we have
$\text{BF}+\text{CF}=\text{EC}+\text{CF}$
$\text{BC}=\text{EF}$
In $\triangle\text{ABC}$ and $\triangle\text{DEF},$
$\angle\text{A}=\angle\text{D}=90^{\circ}$
$\text{BA}=\text{DE}$
$\triangle\text{ABC}\cong\triangle\text{DEF}$
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