$4\vec {AA'}^2-\vec {BC}^2=4\vec{AB}\cdot\vec{AC}$ in triangle

Question

Consider the triangle $ABC$ and the midpoint $A'$ of the side $[BC]$.Show that $4\vec {AA'}^2-\vec {BC}^2=4\vec{AB}\cdot\vec{AC}$.

I have computed that $\vec{AA'}=\frac{\vec{AB}+\vec{AC}}{2}$ but I don't know what to do next. Some tips please?

Answer 1

Hint:

Use cosine law $$\vec{BC}^2=\vec{AB}^2+\vec{AC}^2-2\vec{AB}\cdot\vec{AC}$$ Then \begin{align*} 4\vec{AA'}^2-\vec{BC}^2&=4\left(\frac{\vec{AB}+\vec{AC}}2\right)^2-\vec{BC}^2\\ &=\vec{AB}^2+2\vec{AB}\cdot\vec{AC}+\vec{AC}^2-\left(\vec{AB}^2+\vec{AC}^2-2\vec{AB}\cdot\vec{AC}\right)\\ &=4\vec{AB}\cdot\vec{AC} \end{align*}

Answer 2

$$ \begin{aligned} 4\vec{AA'}^{2}-\vec{BC}^{2}&=4\left(\vec{AA'}+\frac{\vec{BC}}{2}\right)\left(\vec{AA'}-\frac{\vec{BC}}{2}\right)\\ \\ &=4\vec{AC}\cdot\vec{AB} \end{aligned} $$