9-7. Given trapezoid $A B C D$. On its lateral side $C D$, a point $M$ is chosen such that $C M / M D=4 / 3$. It turns out that segment $B M$ divides diagonal $A C$ into two segments, the ratio of whose lengths is also $4 / 3$. What values can the ratio $A D / B C$ take? If necessary, round the answer to 0.01 or write the answer as a common fraction.
Answer. $7 / 12 \approx 0.58$.

Solution. Let segment $B M$ intersect segment $A C$ at point $N$. If $C N / N A = 4 / 3$, then $C N / N A = 4 / 3 = C M / M D$, from which, by the converse of Thales' theorem (or from the similarity of triangles $C N D$ and $C A D$), lines $M N$ and $A D$ are parallel; but this is impossible, because then lines $B C$ and $M N$ would also be parallel, but they intersect at point $B$. Therefore, $N A / N C = 4 / 3$.

![](https://cdn.mathpix.com/cropped/2024_05_06_1d29d5cb430346a58eb5g-23.jpg?height=546&width=1022&top_left_y=1703&top_left_x=517)

Extend segment $B M$ to intersect line $A D$ at point $K$. Triangles $B C N$ and $K A N$ are similar by two angles, from which $K A / B C = N A / N C = 4 / 3$. Triangles $B C M$ and $K D M$ are also similar by two angles, from which $K D / B C = M D / M C = 3 / 4$. It remains to note that

$$
\frac{A D}{B C}=\frac{K A - K D}{B C}=\frac{K A}{B C}-\frac{K D}{B C}=\frac{4}{3}-\frac{3}{4}=\frac{7}{12} \approx 0.58 .
$$