In figure ,`BA _|_ AC,DE _|_DF `such that BA =DE and BF=EC.then

Options

A. `AB=EF`
B. `angleA=angleE`
C. `Delta ABC cong Delta DEF`
D. None

Explanation

Given In figure ,`BA _|_ AC,DE _|_DF `such that BA =DE and BF=EC.
To show `" "Delta ABC cong Delta DEF. `
`Proof Since " " BF=EC `
On adding CF both sides we get
`" "BF+CF=EC+CF`
`In Delta ABC and Delta DEF, " "angleA=angleD=90^(@) " "[because BA_|_ AC and DE _|_ DF]`
`" "BC=EF" "["from Eq.(i)"]`
`and " " BA =DE " "["given"]`
`therefore " "Delta ABC cong Delta DEF " "["by RHS congruence rule"]`