How: $ (\vec b + \vec c)(\vec b - \vec c) = 0 \implies \frac{\vec b + \vec c}{2}.(\vec c - \vec b) = 0 $

Question

I am having difficulty understanding one step of the solution to a question.

question:-
Using vectors, prove that the median to the base of an isosceles triangle is perpendicular to the base.

solution:-

Let ABC be an isosceles triangle in which AB = AC and D is the mid-point of the base BC i.e. AD is median to the base.

TO prove that $$ AD \bot BC \text{, }$$
Let $\vec b$ and $\vec c$ be the position vectors of the points B and C respectively with reference to the point A as the origin, then
$$ P.V. of\;D = \frac{P.V. of \;B + P.V. of \;C}{2} = \frac{\vec b + \vec c}{2}$$
Now $$ AB = AC \\ \implies |\vec{AB}| = |\vec{AC}| \\ \implies |\vec{AB}|^2 = |\vec{AC}|^2 \\ \implies (\vec{AB})^2 = (\vec{AC})^2 \\ \implies (\vec b)^2 = (\vec c)^2 \\ \implies (\vec b)^2 - (\vec c)^2 = 0 \\ \implies (\vec b + \vec c)(\vec b - \vec c) = 0 \\ \implies \frac{\vec b + \vec c}{2}.(\vec c - \vec b) = 0 \\ \implies \vec{AD}.\vec{BC} = 0 \text{ _________________ (since } \vec{BC} = P.V. of\; C - P.V. of\; B = \vec c - \vec b) \\ \implies \vec{AD}\text{ and }\vec{BC} \text{ are at right angles} \\ \implies AD \bot BC \text{ (proved)}$$
I am not able to understand how $(\vec b + \vec c)(\vec b - \vec c) = 0$ implies $\frac{\vec b + \vec c}{2}.(\vec c - \vec b) = 0. $.

Please forgive my formatting of the math code. I am new to math.stack exchange.

Answer 1

If $xy=0$, then also $\frac{x}{2}\cdot -y=0$. Even if done coordinate-wise inside a vector.

Answer 2

Note that

$$\frac{(\vec{b} + \vec{c})(\vec{c} - \vec{b})}{2} = -\frac{(\vec{b} + \vec{c})(\vec{b} - \vec{c})}{2}$$