2024 AMC 10B Problem 10

Because $AD \parallel BC$, triangles $AFE$ and $CFB$ have equal alternate angles at $A, C$ and at $E, B$, so they are similar. The ratio of corresponding sides is $\frac{AE}{CB} = \frac{1}{2}$, since $AE = \frac12 AD = \frac12 BC$.

Set $[AFE] = 1$. Then $[CFB] = 2^2 \cdot 1 = 4$. The similarity also gives $\frac{AF}{FC} = \frac12$.

Triangles $ABF$ and $CBF$ share the vertex $B$ and have bases $AF$ and $FC$ on the same line, so $[ABF] = \frac12 [CBF] = 2$. Hence $[ABC] = 2 + 4 = 6$.

The diagonal $AC$ splits the parallelogram into two equal halves, so $[ACD] = [ABC] = 6$. Quadrilateral $CDEF$ is triangle $ACD$ with triangle $AFE$ removed:
$$[CDEF] = 6 - 1 = 5.$$

The ratio is $[CDEF] : [CFB] = 5 : 4$.

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

A midpoint on one side of a parallelogram creates a $1:2$ similarity with the opposite side, which fixes both the area ratio ($1:4$) and the way the diagonal is cut ($1:2$). From there, "same height, proportional bases" and "a diagonal halves a parallelogram" convert every region into multiples of one unit. Assigning a convenient unit area and propagating ratios is faster than computing any real lengths.

Coordinates: let $ABCD$ be the unit square with $A = (0,1)$, $B = (1,1)$, $C = (1,0)$, $D = (0,0)$, so $E = (0, \tfrac12)$. Lines $EB$: $y = \tfrac12 x + \tfrac12$ and $AC$: $y = 1 - x$ meet at $F = (\tfrac13, \tfrac23)$. Then $[CFB] = \tfrac12 \cdot 1 \cdot \tfrac23 = \tfrac13$ and $[CDEF] = [ACD] - [AFE] = \tfrac12 - \tfrac1{12} = \tfrac5{12}$, ratio $\tfrac{5}{12} : \tfrac13 = 5:4$.

• Using a 1:2 length ratio as a 1:2 area ratio; the areas of similar triangles scale by the square, so [CFB] is four times [AFE].
• Computing the ratio of $CDEF$ to the whole parallelogram, or to triangle $ABE$, instead of to triangle $CFB$.