Vector geometry proof (triangle)

Question

Consider the triangle $ABC$, with point $D$ midpoint of side $AC$, and $E$ midpoint of segment $BD$. Show that $\vec{CE}=\frac{1}{4}\vec{CA}-\frac{1}{2}\vec{BC}.$

What I have tried: since $\vec{CE}=\vec{CD}+\vec{DE}$ I tried to to reformulate $\vec{CD}$ and $\vec{DE}$ in terms of $\vec{CA}$ and $\vec{BC}$. Now, $\vec{CD}=\frac{1}{2}\vec{CA}$ but for $\vec{DE}$ apart from the fact that it is half of $\vec{DB}$ I don't see how to connect it to $\vec{BC}$. I would appreciate an hint about how to proceed with this approach or how I could attack this problem in another way altogether.

Thanks.

Answer 1

Let's consider point $B$ as reference. Then,
$\vec{CE} = \vec{BE} - \vec{BC} = \frac12 \vec{BD} - \vec{BC}$.

Now (considering point $C$ as reference)
$\vec{BD} = \vec{CD} - \vec{CB} = \vec{CD}+\vec{BC} = \frac12\vec{CA} +\vec{BC}$.

So,

$\vec{CE} = \frac14\vec{CA} + \frac12\vec{BC} - \vec{BC} = \frac14\vec{CA}-\frac12\vec{BC}$

Answer 2

Hint 1: If the origin of the plane where the triangle lies is $O$, then $\vec {OX}$ is the position vector, p.v. of point $X$ in the plane, and like you have used, the triangle law of vector addition gives for two points $A,B$, we have $$\vec{OA} + \vec{AB} = \vec{OB} \\ \implies \vec{AB} = p.v(B) - p.v(A)$$

Hint 2: The midpoint of two points $A,B$ identified by their position vectors $(p.v(A)=\vec{OA}, p.v(B)=\vec{OB})$ is given by $C$, such that $$p.v(C) = \dfrac{p.v(A)+p.v(B)}2$$

Note that $p.v(A)=\vec{OA}=a_x \hat{i} + a_y \hat{j}$ (for $2$ dimensions) is just another vector.

Answer 3

Assume the triple $\,(A,B,C)\,$ as a barycentric coordinate system of the plane.

We have, using barycentric coordinates,

$$D=\frac12(A+C) \qquad \text{and}\qquad E=\frac12(B+D)=\frac12 B+\frac14(A+C). $$

Therefore

\begin{align} E-C &= \frac12(B-C)+\frac14(A-C) + \frac14(C-C)=\\[2ex] &=\frac14(A-C)-\frac12(C-B), \end{align}

i.e. our thesis.