In the quadrilateral $A B C D$ given in the figure, we have $A B=5, B C=17$, $C D=5$ and $D A=9$. Determine $D B$, knowing that its measure is an integer.

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-034.jpg?height=214&width=440&top_left_y=2480&top_left_x=1348)
Remember that any side of a triangle is greater than the difference and less than the sum of the other two sides. In triangle $A D B$, we have $A D - A B < B D < A D + A B$ and in triangle $C B D$, we have $B C - C D < B D < B C + C D$.

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-118.jpg?height=248&width=508&top_left_y=821&top_left_x=1325)

Substituting the known values, we get $9 - 5 < B D < 5 + 9$ and $17 - 5 < B D < 17 + 5$, that is,

$$
4 < B D < 14 \text{ and } \quad 12 < B D < 22
$$

From these two inequalities, we conclude that $12 < B D < 14$ and, since $B D$ is an integer, we can only have $B D = 13$.