In a triangle $ABC$,$D$ is the mid-point of side $AC$ such that $BD = \frac{1}{2} AC$. Show that $\angle ABC$ is a right angle.

(N/A) We have to prove that $\angle ABC = 90^{\circ}$.
Since $D$ is the mid-point of $AC$,we have $AD = DC$.
Also,it is given that $BD = \frac{1}{2} AC$.
Since $AC = AD + DC = 2AD$,we have $BD = AD$.
Therefore,$BD = AD = DC$.
In $\Delta ABD$,since $BD = AD$,the angles opposite to equal sides are equal,so $\angle 1 = \angle 2$. $(1)$
In $\Delta BCD$,since $BD = DC$,the angles opposite to equal sides are equal,so $\angle 3 = \angle 4$. $(2)$
In $\Delta ABC$,the sum of angles is $180^{\circ}$:
$\angle A + \angle ABC + \angle C = 180^{\circ}$
$\angle 1 + (\angle 2 + \angle 3) + \angle 4 = 180^{\circ}$
Substituting $\angle 1 = \angle 2$ and $\angle 4 = \angle 3$ from $(1)$ and $(2)$:
$\angle 2 + (\angle 2 + \angle 3) + \angle 3 = 180^{\circ}$
$2(\angle 2 + \angle 3) = 180^{\circ}$
$\angle 2 + \angle 3 = 90^{\circ}$
Since $\angle ABC = \angle 2 + \angle 3$,we have $\angle ABC = 90^{\circ}$.
Hence proved.