2024 AMC 10B Problem 10

The shape can be any parallelogram, so work with areas as multiples of one small piece.

1. Find two matching triangles. Side $AD$ is parallel to side $BC$. So triangle $AFE$ and triangle $CFB$ have the same angles: the angles at $F$ are vertical angles, and the angles at $A$ and $C$ are equal because of the parallel lines. One triangle is a scaled copy of the other.
2. Find the scale. $AE$ is half of $AD$, and $AD = BC$, so $AE$ is half of $CB$. Triangle $AFE$ is a half-size copy of triangle $CFB$.
3. Area of the copy. Halving every length makes the area $\frac12 \cdot \frac12 = \frac14$ as big. Call the area of $AFE$ equal to $1$. Then the area of $CFB$ is $4$.
4. How $F$ cuts the diagonal. The half-size scale also means $AF$ is half of $FC$.
5. Triangle $ABF$. Triangles $ABF$ and $CBF$ both have their top corner at $B$, and their bases $AF$ and $FC$ lie on the same line $AC$. So they have the same height, and $AF$ being half of $FC$ makes the area of $ABF$ half of $4$, which is $2$.
6. Half the parallelogram. Triangle $ABC$ is made of $ABF$ and $CBF$, so its area is $2 + 4 = 6$. The diagonal $AC$ cuts the parallelogram into two equal halves, so triangle $ACD$ also has area $6$.
7. The quadrilateral. $CDEF$ is triangle $ACD$ with the small triangle $AFE$ cut off:
$$6 - 1 = 5.$$
8. So the ratio of $CDEF$ to $CFB$ is $5 : 4$.

The answer is $\boxed{\textbf{(A)}\ 5:4}$.

A midpoint on one side of a parallelogram creates a half-size copy of a triangle on the opposite side. That one fact gives both an area ratio ($1$ to $4$) and how the diagonal is split ($1$ to $2$). After that, "same height, so area follows the base" and "a diagonal cuts a parallelogram in half" turn every region into a multiple of one small unit.

Since the answer is the same for every parallelogram, use a square with side $1$ and coordinates. Put $D = (0, 0)$, $C = (1, 0)$, $B = (1, 1)$, $A = (0, 1)$, so $E = (0, \tfrac12)$.

1. Line $EB$ rises $\frac12$ over a run of $1$: $y = \frac12 x + \frac12$.
2. Line $AC$ goes down from $(0,1)$ to $(1,0)$: $y = 1 - x$.
3. They meet where $\frac12 x + \frac12 = 1 - x$, so $x = \frac13$ and $y = \frac23$. That is $F$.
4. Triangle $CFB$ has base $CB = 1$ on the line $x = 1$, and $F$ is $1 - \frac13 = \frac23$ away from it. Area $= \frac12 \cdot 1 \cdot \frac23 = \frac13$.
5. Triangle $ACD$ has area $\frac12$. Triangle $AFE$ has base $AE = \frac12$ on the line $x = 0$ and height $\frac13$, so area $\frac1{12}$. Then $CDEF$ has area $\frac12 - \frac1{12} = \frac5{12}$.
6. The ratio is $\frac5{12} : \frac13 = 5 : 4$.