Prove that the distance between the center of a circle and every point of a chord is less or equal to the radius. What is stated in the title seems rather obvious.
But is there a geometrical proof of that fact ?
 A: The longest distance from the center to the circumference is the radius (since the diameter is the longest chord). So this means that the distance between the center and a point on a chord couldn't exceed the length of the radius.
A: I give a geometric proof below...
Definitions:
Let the following diagram depict a circle with center $C$ and radius $r$, and let the line segment $\overline{PP'}$ be a chord with endpoints $P$ and $P'$ on the circle. Furthermore, let $\overline{CQ}$ be a line segment whose endpoints are the center $C$ and some arbitrary point $Q$ along the chord. And finally, let $\overline{CT}$ be a line segment whose endpoints are the center $C$ and the point $T$ along the chord such that $\angle CTP = \frac{\pi}{2}$. In other words, $\overline{CT}$ is a line segment that passes through the center $C$ and is perpendicular to $\overline{PP'}$.

For simplicity, I denote the lengths of line segments as follows:
$ |\overline{CP}| = |\overline{CP'}| = r $
$ |\overline{CQ}| = x $
$ |\overline{CT}| = h $
$ |\overline{PQ}| = a $
$ |\overline{QP'}| = b $
$ |\overline{PP'}| = |\overline{PQ}| + |\overline{QP'}| = a+b $
Proof:
We now prove the distance $x$ between the center $C$ of the circle and any point $Q$ of the chord is less than or equal to the radius $r$ as follows:
Note the point $T$ bisects the chord $\overline{PP'}$. You can find proof of this fact here and here. As a result, we have the following
$|\overline{PT}| = \frac{a + b}{2}$
$|\overline{QT}| = \frac{a+b}{2} - a = \frac{b-a}{2}$
and $\triangle CTP$ is a right triangle satisfying the Pythagorean theorem $ r^2 = h^2 + (\frac{a+b}{2})^2 $ which implies $ h^2 = r^2 - (\frac{a+b}{2})^2 $. We now solve for $x$ by noting $\triangle CTQ$ is a right triangle satisfying the Pythagorean theorem
$$x^2 = h^2 + \Big(\frac{b-a}{2}\Big)^2 = r^2 - \Big(\frac{a+b}{2}\Big)^2 + \Big(\frac{b-a}{2}\Big)^2 = r^2 - ab$$
Now, consider that
$r^2 \le r^2 $
$\Leftrightarrow r^2 - ab \le r^2 $
So, by substitution with $x^2 = r^2 - ab$ we have
$ x^2 \le r^2$
$ \Leftrightarrow \sqrt{x^2} \le \sqrt{r^2}$
$ \Leftrightarrow x \le r$
Therefore, the distance $x$ between the center $C$ of the circle and any point $Q$ along a chord is always less than or equal to the radius $r$.
