Inequalities in triangles

Question

Let $D$ be the midpoint of the side $BC$ of triangle $ABC$. Prove that if $AD > BD$ then angle $A$ is acute; else if $AD < BD$, then angle $A$ is obtuse.

Answer 1

$$ |AD|>|BD|=|CD| $$

implies $$ \angle (BAD) < \angle B \quad \mbox{and} \quad \angle (DAC) < \angle C $$ Summing them up you have $$ \angle A < \angle B + \angle C $$ Hence $A$ acute. The other direction is similar.