Is the sum of the lengths of the tangents joining the end points of an arc to a point outside the circle greater than the length of the arc

Question

My book was proving $\lim_{x \to 0}\frac{\sin x}{x}$, and it dished out this line out of nowhere:

$$\text{chord}\ AB<\text{arc}\ APB<AC+BC$$

Now, I can get behind $\text{chord}\ AB<\text{arc}\ APB$. However, how does my book know that $\text{arc}\ APB<AC+BC$?

More info: The radius $OP$ bisects the chord $AB$

Answer 1

However, how does my book know that $\;\text{arc}\ APB<AC+BC ?$

Noting that $OAC$ and $OBC$ are congruent triangles (due to the symmetry of $AC$ and $BC$ being tangents to the circle from a common external point), \begin{aligned} \text{arc } APB&=\frac2{OA}\times\text{area of sector } OAPB \\&< \frac2{OA}\times\text{area of kite } OACB\\&= \frac2{OA}\times OA\times AC\\&\text{(since the tangent $AC$ is perpendicular to the radius $OA$)}\\&=AC+BC.\quad\textit{(shown)} \end{aligned}