Find area of triangle ABD.

Question

Given in $\triangle ABC,~AD$ is the angle bisector of $\angle A $. If area of $\triangle ABC = X$ , prove that area of $\triangle ABD = \dfrac{Xc}{(c+b)}.$

$a=BC$
$b=AC$
$c=AB$

Answer 1

Ratio of base segments on $ BC = \dfrac {c}{b}. $ Since Area = base $\cdot$ height,

fraction of area is (componendo/dividendo proportioning) $$ \frac{c X}{c+b}$$

Answer 2

hint: $\dfrac{\text{area}(\triangle ABD)}{\text{area}(\triangle ABC)}= \dfrac{BD}{BC}= \dfrac{BD}{BD+DC}= \dfrac{\dfrac{BD}{DC}}{1+\dfrac{BD}{DC}}= \dfrac{\dfrac{c}{b}}{1+\dfrac{c}{b}}=...$

Answer 3

$$\frac {c}{b} = \frac {BD}{DC} $$ (by angle bisector theorem)
$$\implies \frac {b+c}{c} = \frac {BC}{BD} = \frac {a}{BD}$$ (after doing reciprocal and then adding 1 to both sides)
$$\implies \frac{BD}{a} = \frac {c}{b+c} $$ $$\implies ar(\triangle ABD) = \frac {BD}{a} * ar(\triangle ABC) $$

$$ = \frac {Xc}{b+c} $$

Hence, proved.