If in a `!ABC` ,CD is the bisector of `angleACB,` then CD =

Options

A. `(a+b)/(2ab)cosC/2`
B. `(a+b)/(ab)cosC/2`
C. `(2ab)/(a+b)cosC/2`
D. `(bsinA)/(sin(B+C/2))`

Explanation

Since AD is the bisector of `angleC` of `Delta` ABC.
`therefore(AD)/(AB)=b/a`
`rArr(AD+DB)/(DB)=(b+a)/arArrc/(DB)=(a+b)/arArrDB=(ac)/(a+b)` …(i)
Also, in `Delta` DCB, we have
`(sinC/2)/(DB)=(sinB)/(CD)rArrDB=(CDsinC/2)/(sinB)` …(ii)

From (i) and (ii), we have
`(ac)/(a+b)=(CDsinC/2)/(sinB)`
`rArrCD=(acsinB)/((a+b)sinC/2)`
`rArrCD=(absinC)/((a+b)sinC/2)rArrCD=(2ab)/(a+b)cosC/2`