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Question
In the given figure, BA ⊥ AC, DE ⊥ DF, such that, BA = DE, BF = EC. Then, by which congruence rule is △ABC ≅ △DEF?
In the given figure, BA ⊥ AC, DE ⊥ DF, such that, BA = DE, BF = EC. Then, by which congruence rule is △ABC ≅ △DEF?
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Solution
The correct option is D RHS
Given, BA ⊥ AC, DE ⊥ DF
and, BA = DE, BF = EC
From the figure, we see that
BC = BF + FC and FE = FC + EC.
Since FC is common in both BC and FE, and BF = EC (given), we must have, BC = FE.
Now, in △ABC and △DEF,
BC = FE (proved)
AB = DE (given)
∠ BAC = ∠EDF = 90∘ (given)
Therefore, △ABC≅△DEF
(by RHS congruence condition)
Given, BA ⊥ AC, DE ⊥ DF
and, BA = DE, BF = EC
From the figure, we see that
BC = BF + FC and FE = FC + EC.
Since FC is common in both BC and FE, and BF = EC (given), we must have, 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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