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Consider the triangle $\Delta ABC$, which $D$ is the midpoint of segment $BC$, and let the point G be defined such that $(GD)= \frac{1}{3}(AD)$. Assuming that $z_A, z_B, z_C$ are the complex numbers representing the points $(A, B, C)$:

a. Find the complex number $z_G$ that represents the point $G$

b. Show that $(CG)= \frac{2}{3}(CF)$ and that $F$ is the midpoint of the segment $(AB)$

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How would you go about solving this? I would apply the distance formula but I am not given any actual complex number. I know that a complex number can be represented as a vector connected to origin.

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Let $z_1, z_2, z_3$ be the points A,B,C. Then it is clear that $$D=\frac{z_2+z_3}{2}$$

The parametric equation of the line from $D$ to $A$ is $$\gamma(t)=\frac{z_2+z_3}{2}+t (z_1 -\frac{z_2+z_3}{2})$$

Therefore $G=\gamma( \frac{1}{3} )$ which you can simplify.

For the second part take the midpoint of AB and repeat the calculation. If you get exactly the same answer then you are done.

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