Question

In the given figure, $BA\perp AC,DE\perp DF,$ such that, $BA=DE,BF=EC.$ Then, $△ABC\cong △DEF$ by which congruence rule?

• A. ASA
• B. SAS
• C. SSS
• D. RHS

Answer 1

The correct option is D RHS

$BC=BF+FC$ and $FE=FC+EC$

Since, $FC$ is common to both sides, and $BF=EC$ [given]

Therefore, $BC=FE$

Now, in $△ABC$ and $△DEF,$

$BC=FE$ [proved]

$AB=DE$ [given]

$\angle BAC=\angle EDF={90}^{\circ }$ [given]

Therefore, $△ABC\cong △DEF$ [by RHS congruence condition]