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.
Picture:
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.
Original question was: May I ask for advice solving this task? Thank you in advance!
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!