The vertices of a triangle are A (-1, -7 ), B ( 5, 1 ) and C (1,4). If the internal angle bisector of $\angle$B meets the side AC in D, then find the length of BD. 


Solution: Let BD be the bisector of  $\angle$ABC. Then, AD : DC = AB : BC
and AB = ${\sqrt{{\left(5 + 1\right)}^2 + \left(1 + 7\right)}}^2 = 10$                                                   
        BC =  ${\sqrt{{\left(5 - 1\right)}^2 + \left(1 - 4\right)}}^2 = 5$
∴ AD : DC = 2 : 1

By section formula, D $\equiv$( 1/3, 1/3)
Distance BD $\sqrt{{\left(5-\frac{1}{3}\right)}^2 + {\left(1-\frac{1}{3}\right)}^2} = \frac{10\sqrt{2}}{3}$.


How did they come up with the first conclusion? AD:DC=AB:BC