Taking refrence to this question
One user is answering that -
In triangle ABC if $\vec{AA_1}$ median then
$\vec{AA_1} = \frac 12 (\vec{AB}+\vec{AC})$
I want to know Is this really a property?
If yes then what is its proof? Because I spent so much time to search but find nothing.
Let $\vec{AK}=2\vec{AA_1}$ Then $ABKC$ are parallelogram. Thus $$2\vec{AA_1}=\vec{AK}=\vec{AB}+\vec{BK}=\vec{AB}+\vec{AC}$$ Then $$\vec{AA_1}=\frac{\vec{AB}+\vec{AC}}2$$
Addition:
In generally:
If $$\frac{AM}{MB}=\frac mn$$ then