Incircle of ` A B C` touches the sides BC, CA and AB at D, E and F, respectively. Let `r_1` be the radius of incircle of ` B D Fdot` Then prove that `r_1=1/2((s-b)sinB)/((1+sin(B/2)))`

Explanation

`angle DIF = pi -B`
Now, `FD = 2FP = 2r sin ((pi -B)/(2)) = 2r cos.(B)/(2)`
Also, `BD = BF = s - b`
Now, in -radius of `DeltaBDF` is `r_(1)`
`:. r_(1) ("Area of "DeltaBDF)/("Semiperimeter of " DeltaBDF)`
`= ((1)/(2) (s-b)^(2) sinB)/((1)/(2) (2(s-b) + 2r cos.(B)/(2)))`
`= ((s-b) sin B)/(2(1 + tan.(B)/(2) cos.(B)/(2))) " " ( :' r = (s-b) tan.(B)/(2))`
`= ((s-b) sin B)/(2(1+ sin.(B)/(2))) ( :' r = (s-b) tan.(B)/(2))`