Question

In Figure $ \mathrm{BA}\perp \mathrm{AC}, \mathrm{DE}\perp \mathrm{DF}$ such that $ \mathrm{BA}=\mathrm{DE}$ and $ \mathrm{BF}=\mathrm{EC}.$ Show that $ \mathrm{\Delta ABC}\cong \mathrm{\Delta DEF}.$

Answer 1

Prove $\mathrm{\Delta ABC}\cong \mathrm{\Delta DEF}.$

Given: $\mathrm{BA}\perp \mathrm{AC},\mathrm{DE}\perp \mathrm{DF},\mathrm{BA}=\mathrm{DE}\mathrm{and}\mathrm{BF}=\mathrm{EC}$

$BF=EC$ (Given)

Add $\mathrm{FC}$ in $\mathrm{BF}$ and $\mathrm{CE}$ we get,

$$\begin{gathered} \mathrm{BF}+\mathrm{FC}=\mathrm{FC}+\mathrm{CE} \\ \Rightarrow \mathrm{BC}=\mathrm{FE}-\left(1\right) \end{gathered}$$

In $∆\mathrm{ABC}$ and $∆\mathrm{DEF}$

$\angle \mathrm{BAC}=\angle \mathrm{FDE}$ (Right angle)

$\mathrm{BC}=\mathrm{FE}$ (From 1)

$\mathrm{AB}=\mathrm{DE}$ (Given)

$\therefore ∆\mathrm{ABC}\cong ∆\mathrm{DEF}$ (by RHS congruency)

Hence proved that $\mathrm{\Delta ABC}\cong \mathrm{\Delta DEF}.$