2005 AMC 10B Problem 23

Write $a = AB$ and $c = DC$. Segment $EF$ joins the midpoints of the legs, so it is parallel to the bases and its length is the average of the bases:
$$EF = \frac{a + c}{2}.$$
It also splits the height $h$ into two equal halves.

Both $ABEF$ and $FECD$ are trapezoids with height $\frac{h}{2}$, so their areas are
$$[ABEF] = \frac{h}{2}\cdot\frac{a + EF}{2}, \qquad [FECD] = \frac{h}{2}\cdot\frac{EF + c}{2}.$$
The condition $[ABEF] = 2[FECD]$ becomes $a + EF = 2(EF + c)$, i.e. $a = EF + 2c$.

Substitute $EF = \frac{a + c}{2}$:
$$a = \frac{a + c}{2} + 2c \quad\Longrightarrow\quad 2a = a + c + 4c \quad\Longrightarrow\quad a = 5c.$$

So $AB/DC = 5$. The answer is $\boxed{\textbf{(C)}\ 5}$.

The midsegment of a trapezoid is the single fact that converts this picture into algebra: it gives the new base length and halves the height at once. Because both sub-trapezoids share the same height, the area condition reduces to a condition on base sums, and everything is linear in $a$ and $c$.

Normalize $DC = 1$ and let $AB = x$. Then $EF = \frac{x+1}{2}$, and the condition $x + \frac{x+1}{2} = 2\left(\frac{x+1}{2} + 1\right)$ simplifies to $x = 5$. Checking the choices in this equation also works quickly.

• Assuming the ratio of areas equals AB/DC directly and answering 2, forgetting that the shared segment EF appears in both areas.
• Using the full height $h$ for one piece and $\frac{h}{2}$ for the other, or forgetting that $EF$ is the average (not the sum) of the bases.