In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$. Prove that $BD = \frac{BC \times AB}{AB + AC}$ and $DC = \frac{BC \times AC}{AB + AC}$.

(N/A) In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$.
By the Angle Bisector Theorem,we have:
$\frac{AB}{AC} = \frac{BD}{DC}$
To find $BD$:
$\frac{AB}{AC} = \frac{BD}{DC}$
Using the property of Componendo,$\frac{AB + AC}{AC} = \frac{BD + DC}{DC} = \frac{BC}{DC}$
This does not directly give $BD$. Let's use $\frac{AB}{AC} = \frac{BD}{DC} \implies \frac{AB}{AB + AC} = \frac{BD}{BD + DC} = \frac{BD}{BC}$
$\therefore BD = \frac{BC \times AB}{AB + AC}$
To find $DC$:
$\frac{AB}{AC} = \frac{BD}{DC} \implies \frac{AC}{AB} = \frac{DC}{BD}$
Using Componendo,$\frac{AC + AB}{AB} = \frac{DC + BD}{BD} = \frac{BC}{BD}$
$\implies \frac{AC}{AC + AB} = \frac{DC}{BC}$
$\therefore DC = \frac{BC \times AC}{AB + AC}$