Question

in the given figure.BA=AC and DE=EF such that BA=DE and BE= DC. prove that AC= EF
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Answer 1

Answer:

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)

AC = EF ( BY CPCT)

Step-by-step explanation:

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