1. In a triangle, two sides $a$ and $b(a \geq b)$ are given. Find the third side if it is known that $a+h_{a} \geq b+h_{b}$ where $h_{a}$ and $h_{b}$ are the heights dropped to $a$ and $b$ respectively.
Solution. Let $\gamma$ be the angle between the sides

![](https://cdn.mathpix.com/cropped/2024_06_05_41f37b35b75ee031cfebg-4.jpg?height=336&width=561&top_left_y=1846&top_left_x=921)
$a$ and $b$; then we have

$$
\begin{aligned}
& a+b \sin \gamma \leq b+a \sin \gamma \\
& (a-b) \sin \gamma \geq(a-b) \\
& (a-b)(\sin \gamma-1) \geq 0 \\
& \sin \gamma-1 \geq 0
\end{aligned}
$$

from which we obtain $\gamma=90^{\circ}$, i.e., $c=\sqrt{a^{2}+b^{2}}$.