# Triangular prism, centroid of a triangle

Let $$ABCA_1B_1C_1$$ be a triangular prism. There is a point $$T\in\Delta ABC$$. Let $$T_1$$ be a centroid of $$\Delta A_1C_1T$$. If the following holds: $$\overrightarrow{A_1T_1}=\frac{1}{3}\overrightarrow{A_1A}+\frac{1}{9}\overrightarrow{A_1B_1}+\frac{4}{9}\overrightarrow{A_1C_1}$$ prove that $$T$$ is the centorid of $$\Delta ABC$$.

My attempt:

I wanted to use the following fact:

Let $$\overline{AT}$$ be a median to $$\overline{BC}$$. Then: $$\overrightarrow{AT}=\frac{\overrightarrow{AB}+\overrightarrow{AC}}{2}=\overrightarrow{AB}+\frac{\overrightarrow{BC}}{2}=\overrightarrow{AC}-\frac{\overrightarrow{BC}}{2}$$ because a median is half of a diagonal of a parallelogram.

I expressed $$\overrightarrow{AT_1}$$ as: $$\overrightarrow{A_1T_1}=\frac{1}{3}\left(\overrightarrow{A_1T}+\overrightarrow{A_1C_1}\right)$$ and $$\frac{1}{9}\overrightarrow{A_1B_1}+\frac{4}{9}\overrightarrow{A_1C_1}$$ as: $$\frac{1}{9}\overrightarrow{A_1B_1}+\frac{4}{9}\overrightarrow{A_1C_1}=\frac{1}{3}\left(\overrightarrow{A_1B_1}+\overrightarrow{A_1C_1}\right)-\frac{2}{9}\overrightarrow{A_1B_1}+\frac{1}{9}\overrightarrow{A_1C_1}$$

I also considered expanding the triangular prism to a quadrilateral prism. Then: $$\overrightarrow{A_1T_1}=\frac{1}{3}\overrightarrow{A_1A_1^{'}},$$ but it was useless.

Update: It was only until this morning I noticed a mistake on the picture that was misleading (thanks to @MichaelRozenberg in the answer) The position of $$T_1$$ was wrong. I don't want any reader to be confused with a chaotic picture, so I replaced it with an accurate one.

Let $$\vec{A_1B_1}=\vec{u},$$ $$\vec{A_1C_1}=\vec{v},$$ $$\vec{A_1A}=\vec{w}$$ and $$\vec{AT}=\alpha\vec{u}+\beta\vec{v}.$$
Thus, $$\vec{A_1T_1}=\frac{1}{3}(\vec{w}+\alpha\vec{u}+\beta\vec{v}+\vec{v})$$ and $$\frac{1}{3}(\vec{w}+\alpha\vec{u}+(1+\beta)\vec{v})=\frac{1}{3}\vec{w}+\frac{1}{9}\vec{u}+\frac{4}{9}\vec{v},$$ which gives $$(\alpha,\beta)=\left(\frac{1}{3},\frac{1}{3}\right)$$ and we are done!