Let $s$ denote the semi-perimeter of a triangle $ABC$ in which $BC = a, CA = b, AB = c$. If a circle touches the sides $BC, CA, AB$ at $D, E, F$ respectively,prove that $BD = s - b$.

(N/A) circle is inscribed in the $\triangle ABC$,which touches the sides $BC, CA$ and $AB$ at $D, E$ and $F$ respectively.
Given,$BC = a, CA = b$ and $AB = c$.
By using the property that tangents drawn from an external point to a circle are equal in length:
$\therefore BD = BF = x$ (let)
$DC = CE = y$ (let)
$AE = AF = z$ (let)
The perimeter of $\triangle ABC = BC + CA + AB = a + b + c$.
Since $2s = a + b + c$,we have:
$(BD + DC) + (CE + EA) + (AF + FB) = a + b + c$
$(x + y) + (y + z) + (z + x) = a + b + c$
$2(x + y + z) = 2s$
$x + y + z = s$
We need to find $BD = x$.
From the equation $x + y + z = s$,we get $x = s - (y + z)$.
Since $b = CA = CE + EA = y + z$,substituting this into the equation:
$x = s - b$
Therefore,$BD = s - b$.
Hence proved.