4. The given figure is an isosceles triangle $ABC$ with base $\overline{AB}$ of length $65 \, \text{cm}$ and legs $\overline{AC}$ and $\overline{BC}$ of length $80 \, \text{cm}$. On the base $\overline{AB}$, a point $D$ is chosen such that the perimeter of triangle $ADC$ is $173 \, \text{cm}$, and the perimeter of triangle $DBC$ is $220 \, \text{cm}$. What are the lengths of segments $\overline{CD}, \overline{AD}$, and $\overline{DB}$?
Another solution.

Let $b$ be the length of the legs of triangle $ABC$, $x$ be the length of segment $\overline{AD}$, $y$ be the length of segment $\overline{DB}$, and $z$ be the length of segment $\overline{CD}$. From the given, the base $\overline{AB}$ has a length of $x+y=65 \text{ cm}$.

Sketch:

![](https://cdn.mathpix.com/cropped/2024_05_30_5715e7b36c946fa665d7g-06.jpg?height=539&width=425&top_left_y=987&top_left_x=290)

Let's calculate the perimeter of triangle $ABC$.

$O_{ABC}=|AB|+|BC|+|CA|$

$O_{ABC}=65+80+80$

$O_{ABC}=225 \text{ cm}$

1 POINT

It is given that

$O_{ABC}=x+y+2b$ or $225=x+y+2b$

$O_{ADC}=x+b+z$ or $173=x+b+z$

$O_{DBC}=y+b+z$ or $220=y+b+z$

1 POINT

If we add the perimeters of triangle $ADC$ and triangle $DBC$, we get

$173+220=\underbrace{x+y+2b}_{225}+2z$

so it follows that:

$2 \cdot z=173+220-225$

1 POINT

$2 \cdot z=168$

$z=|DC|=84 \text{ cm}$

1 POINT

$x$ is calculated from the perimeter of triangle $ADC$.

$80+x+84=173$

1 POINT

$x=173-(80+84)$

$x=|AD|=9 \text{ cm}$

1 POINT

$y$ is calculated from the known lengths of segments $\overline{AB}$ and $\overline{AD}$.

$y=65-9$

$y=|DB|=56 \text{ cm}$

1 POINT

Note: Points are not deducted for the use of improper mathematical notation or lack of units. For 1 POINT, the sketch must have the vertices of all mentioned triangles correctly labeled (not necessarily in a counterclockwise direction).

TOTAL 10 POINTS