If the lengths of the sides BC, CA and AB of a $\triangle\text{ABC}$ are a, b and c respectively and AD is the bisectore of $\angle\text{A}$ then find the lengths of BD and DC.
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Given $\triangle\text{ABC}$ in which AD, the bisector of $\angle\text{A}$ meets BC in D.
Let = x ⇒ DC = (a - x)
Then by the angle Bisector theorem,
$\frac{\text{BD}}{\text{DC}}=\frac{\text{AB}}{\text{AC}}$
$\Rightarrow\frac{\text{x}}{\text{a}-\text{x}}=\frac{\text{c}}{\text{b}}$
$\Rightarrow\text{xb}=\text{c}(\text{a}-\text{x})$
$\Rightarrow\text{xb}=\text{ac}-\text{xc}$
$\Rightarrow\text{xb}+\text{xc}=\text{ac}$
$\Rightarrow\text{x}=\frac{\text{ac}}{\text{b+c}}$
So, $\text{BD}=\frac{\text{ac}}{\text{b+c}}$
$\text{DC}=\text{a}-\text{x}\Rightarrow\text{DC}=\text{a}-\frac{\text{ac}}{\text{b+c}}$
$=\frac{\text{ab}+\text{ac}-\text{ac}}{\text{b}+\text{c}}=\frac{\text{ab}}{\text{b}+\text{c}}$
Hence, $\text{BD}=\frac{\text{ac}}{\text{b}+\text{c}}$ and $\text{DC}=\frac{\text{ab}}{\text{b}+\text{c}}.$
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