Problem 4. Let the isosceles triangle $ABC$ be given, where $\overline{AB}=\overline{CB}$, with base $\overline{AC}=10 \text{ cm}$. Through the midpoint $D$ of the leg $AB$, a normal $n$ to the leg $AB$ is drawn, intersecting the leg $CB$ at point $E$. If the perimeter of triangle $ABC$ is $40 \text{ cm}$, calculate the perimeter of triangle $AEC$.

![](https://cdn.mathpix.com/cropped/2024_06_05_dc4d34c0a9dd2a3ba572g-06.jpg?height=445&width=358&top_left_y=1758&top_left_x=1109)
Solution. Let $ABC$ be an isosceles triangle ($\overline{AB}=\overline{CB}$), $D$ the midpoint of the leg $AB$, and $n$ the normal to $AB$ through $D$ which intersects the leg $CB$ at point $E$ (see diagram). From $\overline{AC}=10 \text{ cm}$ and $L_{ABC}=40 \text{ cm}$, it follows that the length of each leg is $\overline{AB}=\overline{CB}=(40-10): 2=15 \text{ cm}$. The triangles $ADE$ and $BDE$ are congruent (SAS: $\overline{AD}=\overline{BD}$ from $D$ being the midpoint of the leg $AB$, $\angle ADE=\angle BDE=90^{\circ}$, $\overline{DE}$ is a common side). From this, it follows that $\overline{AE}=\overline{BE}$. Now, the perimeter of triangle $AEC$ is

$$
L_{AEC}=\overline{AE}+\overline{EC}+\overline{AC}=(\overline{BE}+\overline{EC})+\overline{AC}=\overline{BC}+\overline{AC}=15+10=25 \text{ cm}
$$