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Question

In the given figure, BA ⊥ AC, DE ⊥ DF such that BA = DE and BF = EC. Show that Δ ABC Δ DEF


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Solution

According to the question,

BA ⊥ AC, DE ⊥ DF such that BA = DE and BF = EC.

In ΔABC and ΔDEF

BA = DE (given)

BF = EC (given)

∠A = ∠D (both 90°)

BC = BF + FC

EF = EC + FC = BF + FC ( EC = BF)

EF = BC

Hence, ΔABC ΔDEF (by RHS)

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