In the given figure,$BA \perp AC$ and $DE \perp DF$ such that $BA = DE$ and $BF = EC$. Show that $\triangle ABC \cong \triangle DEF$.
(N/A) Given: $BA \perp AC$,$DE \perp DF$,$BA = DE$,and $BF = EC$.
Step $1$: Establish the equality of the hypotenuses.
We are given $BF = EC$.
Adding $FC$ to both sides,we get:
$BF + FC = EC + FC$
$BC = EF$
Step $2$: Compare the two triangles.
In $\triangle ABC$ and $\triangle DEF$:
$1$. $\angle A = \angle D = 90^{\circ}$ (Given as $BA \perp AC$ and $DE \perp DF$)
$2$. $BA = DE$ (Given)
$3$. $BC = EF$ (Hypotenuse,proved above)
Step $3$: Conclusion.
By the $RHS$ (Right angle-Hypotenuse-Side) congruence rule,$\triangle ABC \cong \triangle DEF$.
Step $1$: Establish the equality of the hypotenuses.
We are given $BF = EC$.
Adding $FC$ to both sides,we get:
$BF + FC = EC + FC$
$BC = EF$
Step $2$: Compare the two triangles.
In $\triangle ABC$ and $\triangle DEF$:
$1$. $\angle A = \angle D = 90^{\circ}$ (Given as $BA \perp AC$ and $DE \perp DF$)
$2$. $BA = DE$ (Given)
$3$. $BC = EF$ (Hypotenuse,proved above)
Step $3$: Conclusion.
By the $RHS$ (Right angle-Hypotenuse-Side) congruence rule,$\triangle ABC \cong \triangle DEF$.
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