A straight line cuts side AB of triangle ABC at F and side CA at E, and meets the line BC at a point D outside the segment BC. If AF : FB = 2 : 1 and AE : EC = 2 : 3, what is the ratio BD : DC?

Menelaus' theorem for the transversal $DEF$ gives $(AF/FB)\times (BD/DC)\times (CE/EA)=1$, each ratio taken in the same order round the triangle. Here $AF/FB=2$ and $CE/EA=3/2$, so $2\times (BD/DC)\times 3/2=1$, giving $BD:DC=1$ : 3. Check with A(0, 0), $B(3,0),C(0,5):F=(2,0),E$ = (0, 2), the line $x+y=2$ meets $BC$ at $D$(4.5, −2.5), and $BD:DC=1$ : 3. Using $AE/EC$ instead of $CE/EA$ gives the trap 3 : 4; 3 : 1 is the answer upside down.