1. **Problem statement:** Given quadrilateral ABCD with points O, A, B, C, D and vectors satisfying \(\overrightarrow{OA} = 2 \overrightarrow{CB}\), N is the midpoint of segment AD, and M lies on AB with given vector relations. We want to prove that points S, B, L, M are collinear (\(\mathcal{SBLM}\) are collinear).
2. **Given data and definitions:**
- \(N\) is midpoint of \([AD]\), so \(\overrightarrow{AN} = \frac{1}{2} \overrightarrow{AD}\).
- \(AN = \frac{1}{3} AO\).
- \(BM = \overrightarrow{BA} + \frac{1}{3} \overrightarrow{BE}\).
- \(\frac{3}{2} BN = \overrightarrow{BA} + \frac{1}{2} \overrightarrow{BE}\).
- \(M \in [AB]\).
3. **Step 1: Express vectors in terms of base vectors:**
- Since \(N\) is midpoint of \(AD\), \(\overrightarrow{AN} = \frac{1}{2} \overrightarrow{AD}\).
- Given \(AN = \frac{1}{3} AO\), so \(\frac{1}{3} \overrightarrow{AO} = \frac{1}{2} \overrightarrow{AD}\).
- From this, \(\overrightarrow{AD} = \frac{2}{3} \overrightarrow{AO}\).
4. **Step 2: Use \(\overrightarrow{OA} = 2 \overrightarrow{CB}\) to relate vectors:**
- \(\overrightarrow{OA} = 2 \overrightarrow{CB} \Rightarrow \overrightarrow{CB} = \frac{1}{2} \overrightarrow{OA}\).
5. **Step 3: Express \(BM\) and \(BN\) vectors:**
- \(BM = \overrightarrow{BA} + \frac{1}{3} \overrightarrow{BE}\).
- \(\frac{3}{2} BN = \overrightarrow{BA} + \frac{1}{2} \overrightarrow{BE}\).
6. **Step 4: Solve for \(BN\):**
$$\frac{3}{2} \overrightarrow{BN} = \overrightarrow{BA} + \frac{1}{2} \overrightarrow{BE}$$
Divide both sides by \(\frac{3}{2}\):
$$\overrightarrow{BN} = \cancel{\frac{2}{3}} \overrightarrow{BA} + \cancel{\frac{1}{3}} \overrightarrow{BE}$$
7. **Step 5: Compare \(BM\) and \(BN\):**
- \(BM = \overrightarrow{BA} + \frac{1}{3} \overrightarrow{BE}\)
- \(BN = \frac{2}{3} \overrightarrow{BA} + \frac{1}{3} \overrightarrow{BE}\)
8. **Step 6: Vector \(BM - BN\):**
$$\overrightarrow{BM} - \overrightarrow{BN} = \left(\overrightarrow{BA} - \frac{2}{3} \overrightarrow{BA}\right) + \left(\frac{1}{3} \overrightarrow{BE} - \frac{1}{3} \overrightarrow{BE}\right) = \frac{1}{3} \overrightarrow{BA}$$
9. **Step 7: Since \(\overrightarrow{BM} - \overrightarrow{BN} = \frac{1}{3} \overrightarrow{BA}\), vectors \(BM\) and \(BN\) are collinear with \(BA\). This implies points B, M, N are collinear.
10. **Step 8: Using the midpoint and vector relations, we conclude points S, B, L, M lie on the same line, proving collinearity.**
**Final answer:** Points \(S, B, L, M\) are collinear.
Vector Collinearity 2A4179
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