`DeltaABC` की भुजा AC पर `B_(1)` एक बिंदु है और `B_(1)B` मिला दी जाती है `B_(1)B` के समांतर A से होकर एक रेखा खींची जाती है तो BC से `A_(1)` पर मिलती है और `B_(1)B` के समांतर एक और रेखा खींची जाती है जो C से होकर जाती है और AB से `C_(1)` पर मिलती है, तो

Options

A. `(1)/(C C_(1)) -(1)/(A A_(1)) = (1)/(B B_(1))`
B. `(1)/(C C_(1))+(1)/(A A_(1)) = (1)/(B B_(1))`
C. `(1)/(B B_(1)) -(1)/(A A_(1)) = (1)/(C C_(1))`
D. `(1)/(B B_(1)) +(1)/(A A_(1)) =(1)/(C C_(1))`

Answer

B

Explanation

In `DeltaAA_(1)C ~=DeltaBB_(1)C`
`(BB_(1))/(AA_(1)) = (B_(1)C)/(AC)`………….(i)
`DeltaAC_(1)C ~=DeltaBB_(1)A`
`(BB_(1))/(C C_(1)) = (AB_(1))/(AC)`…………(ii)
Adding eq. (i) and (ii)
`(BB_(1))/(AA_(1)) +(BB_(1))/(C C_(1)) = (B_(1)C)/(AC) +(AB_(1))/(AC)`
`BB_(1)[1/(A A_(1)) + 1/(C C_(1))] = (B_(1)C+AB_(1))/(AC)`
`1/(AA_(1)) + 1/(C C_(1)) = (AC)/(AC) xx 1/(BB_(1))` or `1/(B B_(1)) = 1/(AA_(1)) + 1/(C C_(1))`