4. The line $A D$ to the base $B C$ of the isosceles triangle $A B C$ divides the perimeter of the same triangle into parts of $17 \mathrm{~cm}$ and $11 \mathrm{~cm}$. Find the sides of the triangle if their numerical values are consecutive natural numbers.
Solution. From the condition of the problem, it follows that
a) $\overline{A C}+\overline{C D}=17 ; \overline{A B}+\overline{B D}=11$ or

b) $\overline{A C}+\overline{C D}=11 ; \overline{A B}+\overline{B D}=17$.

a) By adding these two equations, we get $\overline{A C}+\overline{C D}+\overline{A B}+\overline{B D}=28$, i.e.,

$$
2 \overline{A C}+\overline{A B}=28
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_d112afdaa6f4bb49aaeeg-2.jpg?height=423&width=469&top_left_y=901&top_left_x=1015)

Since the lengths of the sides of the triangle are consecutive natural numbers, from the last equation it follows that

$$
2 \overline{A C}+\overline{A C}+1=28 \text { or } 2(\overline{A B}+1)+\overline{A B}=28
$$

where a solution is obtained only in the first case, and that is $\overline{A C}=9 ; \overline{A B}=10$.

b) By adding these two equations, we get $2 \overline{A C}+\overline{A B}=28$ from which, as in case a), we find $\overline{A C}=9 ; \overline{A B}=10$.

## VII Section