1. **State the problem:** We have triangle $ABC$ with medians $\overline{AE}$, $\overline{BF}$, and $\overline{CD}$ intersecting at centroid $G$. Given $GF=8$, $GD=4$, and $AE=33$, find lengths $CD$, $AG$, and $BG$. 2. **Recall properties of centroid and medians:** The centroid $G$ divides each median into a ratio $2:1$, with the longer segment adjacent to the vertex. 3. **Find $CD$:** Since $G$ divides median $CD$ such that $CG:GD=2:1$, and $GD=4$, then $$CD = CG + GD = 2 \times GD + GD = 3 \times GD = 3 \times 4 = 12.$$ 4. **Find $AG$:** $G$ divides median $AE$ in ratio $2:1$, so $$AG = \frac{2}{3} AE = \frac{2}{3} \times 33 = 22.$$ 5. **Find $BG$:** $G$ divides median $BF$ in ratio $2:1$. Given $GF=8$, then $$BF = BG + GF = 2 \times GF + GF = 3 \times GF = 3 \times 8 = 24.$$ Since $BG$ is twice $GF$, $$BG = 2 \times GF = 2 \times 8 = 16.$$ **Final answers:** $$CD = 12, \quad AG = 22, \quad BG = 16.$$