Let $ABCDEF$ be a convex hexagon of area $1$, whose opposite sides are parallel. The lines $AB$, $CD$ and $EF$ meet in pairs to determine the vertices of a triangle. Similarly, the lines $BC$, $DE$ and $FA$ meet in pairs to determine the vertices of another triangle. Show that the area of at least one of these two triangles is at least $3/2$.
1. Let \(EF\) and \(AB\) meet at \(P_1\), \(AB\) and \(CD\) meet at \(Q_1\), and \(CD\) and \(EF\) meet at \(R_1\).
2. Let \(BC\) and \(DE\) meet at \(P_2\), \(DE\) and \(FA\) meet at \(Q_2\), and \(FA\) and \(BC\) meet at \(R_2\).
3. Let \(\angle Q_1P_1R_1 = \alpha\), \(\angle R_1Q_1P_1 = \beta\), and \(\angle P_1R_1Q_1 = \gamma\).
4. Let \(p = \sin \alpha\), \(q = \sin \beta\), and \(r = \sin \gamma\). Without loss of generality, assume that \(|FA| \leq |CD|\). Let \(|FA| = pa\), \(|BC| = qb\), \(|DE| = rc\), \(|CD| = px\), \(|EF| = qy\), and \(|AB| = rz\).
5. By simple angle chasing, we find that \(\triangle P_1AF\), \(\triangle BAR_2\), \(\triangle BQ_1C\), \(\triangle P_2DC\), \(\triangle EDR_1\), \(\triangle EQ_2F\), \(\triangle P_1Q_1R_1\), and \(\triangle P_2Q_2R_2\) are similar.
6. Using the sine rule on \(\triangle P_1AF\):
   \[
   \frac{|FP_1|}{q} = \frac{|P_1A|}{r} = \frac{|FA|}{p} = a \implies |FP_1| = qa \text{ and } |P_1A| = ra
   \]
   Similarly, we find:
   \[
   |BQ_1| = rb, \quad |Q_1C| = pb, \quad |DR_1| = pc, \quad |R_1E| = qc, \quad |CP_2| = qx, \quad |P_2D| = rx, \quad |EQ_2| = ry, \quad |Q_2F| = py, \quad |AR_2| = pz, \quad |R_2B| = qz
   \]
7. Therefore:
   \[
   |Q_1R_1| = p(x + b + c), \quad |R_1P_1| = q(a + y + c), \quad |P_1Q_1| = r(a + b + z)
   \]
   \[
   |Q_2R_2| = p(a + y + z), \quad |R_2P_2| = q(x + b + z), \quad |P_2Q_2| = r(x + y + c)
   \]
8. Using the sine rule on \(\triangle P_1Q_1R_1\):
   \[
   \frac{|Q_1R_1|}{p} = \frac{|R_1P_1|}{q} = \frac{|P_1Q_1|}{r} \implies x + b + c = a + y + c = a + b + z
   \]
   Let \(d = x - a = y - b = z - c\). Then:
   \[
   |Q_1R_1| = p(a + b + c + d), \quad |R_1P_1| = q(a + b + c + d), \quad |P_1Q_1| = r(a + b + c + d)
   \]
   \[
   |Q_2R_2| = p(a + b + c + 2d), \quad |R_2P_2| = q(a + b + c + 2d), \quad |P_2Q_2| = r(a + b + c + 2d)
   \]
9. Since \(|FA| \leq |CD|\), we have \(pa \leq px \implies d = x - a \geq 0\).
10. The area of hexagon \(ABCDEF\) is given by:
    \[
    \text{Area of } ABCDEF = \text{Area of } \triangle P_1Q_1R_1 - \text{Area of } \triangle P_1AF - \text{Area of } \triangle CQ_1B - \text{Area of } \triangle EDR_1
    \]
    \[
    1 = \frac{pqr(a + b + c + d)^2}{2} - \frac{pqra^2}{2} - \frac{pqrb^2}{2} - \frac{pqrc^2}{2}
    \]
    \[
    \implies d^2 + 2d(a + b + c) + 2ab + 2bc + 2ca = \frac{2}{pqr}
    \]
11. We know that:
    \[
    (a - b)^2 + (b - c)^2 + (c - a)^2 \geq 0 \implies a^2 + b^2 + c^2 \geq ab + bc + ca
    \]
    Therefore:
    \[
    \frac{3}{2}d^2 + 3d(a + b + c) + (a^2 + b^2 + c^2) + (2ab + 2bc + 2ca) \geq \frac{3}{pqr}
    \]
    Since \(d \geq 0\), we have:
    \[
    4d^2 + 4d(a + b + c) + (a^2 + b^2 + c^2) + (2ab + 2bc + 2ca) \geq \frac{3}{pqr}
    \]
    \[
    \implies (a + b + c + 2d)^2 \geq \frac{3}{pqr}
    \]
    \[
    \implies \frac{pqr(a + b + c + 2d)^2}{2} \geq \frac{3}{2}
    \]
12. Thus, the area of \(\triangle P_2Q_2R_2\) is:
    \[
    \text{Area of } \triangle P_2Q_2R_2 = \frac{pqr(a + b + c + 2d)^2}{2} \geq \frac{3}{2}
    \]

Therefore, the area of at least one of the two triangles is at least \(\frac{3}{2}\).

The final answer is \( \boxed{ \frac{3}{2} } \)