Suppose $O$ is the centre of the circumscribing circle of triangle $ABC$ and $H$ is its orthocentre. Prove that vector $OH$ is equal to the sum of the vectors $OA$, $OB$ and $OC$.
An answer I found on the Internet.
Let's start with the following observation. For any vectors $U$ and $V$, if $\lvert U\rvert = \lvert V\rvert$ then $U+V$ is perpendicular to $U-V$. We show this using the dot product:
$$(U+V)\cdot(U-V) = \lvert U\rvert^2 - U\cdot V + V\cdot U - \lvert V\rvert^2 = 0$$
Now for simplicity we will use different notation from the first proof. Let the vectors from the center of the circumscribed circle to the vertices of the triangle be $A$, $B$ and $C$. Note that $\lvert A\rvert = \lvert B\rvert = \lvert C\rvert$ and vectors $A-B$, $B-C$ and $C-A$ represent the three sides of the triangle. We want to show that the orthocenter is the point $A+B+C$.
Now construct the orthocenter as the intersection of two altitudes of the triangle. Since $B+C$ is perpendicular to $B-C$ the altitude from $A$ is on the line with parametric equation $$L_1 = A + a(B+C),$$ where $a$ is the parameter. Also, since $A+B$ is perpendicular to $A-B$ the altitude from $C$ is on the line with parametric equation $$L_2 = C + b(A+B),$$ where $b$ is the parameter. When $L_1$ and $L_2$ intersect, $$ A + a(B+C) = C + b(A+B)$$ and $$ a(B+C) = C - A + b(A+B).$$
Now take the dot product of both sides with $A-B$. Note that this causes $b$ to drop out, which allows us to solve for $a$: $$ a(B+C)\cdot (A-B) = (C-A)\cdot (A-B)$$ $$a(B\cdot A - \lvert B\rvert^2 + C\cdot A - C\cdot B) = C\cdot A - C\cdot B - \lvert A\rvert ^2 + A\cdot B.$$
Since $\lvert B\rvert = \lvert A\rvert$ the quantity in the parentheses is equal to the right hand side. It can be shown that this quantity is non-zero, so we can divide by it to arrive at $$ a = 1.$$
The orthocenter is the value of $L_1$ when $a = 1$. Therefore, the orthocenter is the point $A+B+C$ as was to be shown, so the proof is complete.
However, I understood EVERY WORD OF it except this:
Since $B+C$ is perpendicular to $B-C$ the altitude from $A$ is on the line with parametric equation $$ L_1 = A + a(B+C).$$
Would someone guide me please? What do they mean by this line?