How to prove that in any triangle $ABC$, there exists a point $D$ on the longest side of the triangle, $\overline{BC}$, such that $\overline{AD} \perp \overline{BC}$, using only geometry.
I tried proving this by contradiction, by assuming that for every point $D$ on $\overline{BC}$, $\angle ADB>90\unicode{xB0}$ or $\angle ADB<90\unicode{xB0}$. Let $D$ be any point on $\overline{BC}$. For the first case, $\angle ADB>90\unicode{xB0}$, I've only managed to show that $\angle DAB + \angle B < 90\unicode{xB0}$. I wish to show somehow show that $\angle A < \angle C$ or $\angle A < \angle B$, then I can show that $\overline{BC}$ isn't the longest side of the triangle. Thanks in advance
Since $BC$ is the longest side, we have $B,C<\frac{\pi}{2}$. Therefore, you can always draw a line $AD$ such that $\angle CAD=\frac{\pi}{2}-C$ and $\angle BAD=\frac{\pi}{2}-B$ like the following:
Since the sum of angles in a triangle must be $\pi$, we have $\angle CDA = \frac{\pi}{2}$, i.e., $AD\perp BC$