Question

in ∆ABC , AD the bisector of angle A meets BC at D. If BC=a, AC=b ,AB=c
then BD-DC​

Answer 1

$\textbf{Given:}$

$\text{In \triangle ABC, AD is the bisector of \angle{A} }$

$\text{and BC=a, AC=b, AB=c}$

$\textbf{To find:}$

$\text{BD-DC}$

$\textbf{Solution:}$

$\textbf{Angle bisector theorem:}$

$\textbf{When vertical angle of a triangle is bisected,}$

$\textbf{the bisector divides the base into two segments}$

$\textbf{which have the ratio as the order of other two sides}$

$\text{Since AD is the bisector of \angle{A} , we get}$

$\text{By angle bisector theorem}$

$\dfrac{BD}{DC}=\dfrac{AB}{AC}$

$\dfrac{BD}{DC}=\dfrac{c}{b}$

$\implies\,BD=ck\;\;\text{and}\;\;DC=bk$

$\text{But}\;BC=BD+DC$

$\implies\,a=ck+bk$

$\implies\,a=(b+c)k$

$\implies\,k=\dfrac{a}{b+c}$

$\text{Now,}$

$BD-DC=ck-bk$

$BD-DC=(c-b)k$

$BD-DC=(c-b)\dfrac{a}{b+c}$

$\implies\boxed{\bf\,BD-DC=\dfrac{a(c-b)}{b+c}}$

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