Finding $\vec{DE}$ in a triangle $ABC$.

Question

Have a triangle with vectors $A,B,C$. Then have $D$ as the middle point of $\overline{AB}$ and $E$ as the middle point of $\overline{AC}$:

Demonstrate that $\vec{DE} = \frac{1}{2}\vec{BC}$.

Observe that $$\frac{1}{2}\vec{BC} = \frac{1}{2}\vec{AB} - \frac{1}{2}\vec{AC}$$

From the hypothesis, we have that

$$\frac{1}{2}\vec{AB} = \vec{AD}$$

and

$$\frac{1}{2}\vec{AC} = \vec{AE}$$

So:

$$\frac{1}{2}\vec{BC} = \vec{AD} - \vec{AE}$$

$$\frac{1}{2}\vec{BC} = \vec{ED}$$

Hmm... I'm not sure what to say. Technically, this is not what I wanted to demonstrate (I wanted $\vec{DE}$, not $\vec{ED}$).

However, $\vec{ED}$ clearly has the same length as $\vec{DE}$, so it kind of works out at the end...

... What do you think? Is this proof good for this, or did I make some mistake somewhere?

Answer 1

In your diagram you took direction of $\vec{DE}$ opposite to $\vec{BC}$. Hence, you get the result. Your result is correct.

Answer 2

if you are defining $$ \vec{BC} = \vec{AB} - \vec{AC} $$ then, $$ \vec{AD} - \vec{AE} = \vec{DE} $$ and not $$ \vec{ED} $$

The actual problem is with first statement, it should be

$$ \vec{BC} = \vec{AC} - \vec{AB} $$