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Question
In the given figure, BA⊥AC, DE⊥DF, such that, BA=DE, BF=EC. Then, △ABC≅△DEF by which congruence rule?
In the given figure, BA⊥AC, DE⊥DF, such that, BA=DE, BF=EC. Then, △ABC≅△DEF by which congruence rule?
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Solution
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]
∠BAC=∠EDF=90∘ [given]
Therefore, △ABC≅△DEF [by RHS congruence condition]
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]
∠BAC=∠EDF=90∘ [given]
Therefore, △ABC≅△DEF [by RHS congruence condition]
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