Is the Incenter always “below” the middle point of an angle bisector segment in a triangle?

Let $ABC$ be any triangle. Segment $\overline{CD}$ bisects angle $\angle ACB$, and segment $\overline{EB}$ bisects angle $\angle ABC$. These segments meet at the point $I$ (the incenter of the triangle). Prove that the middle point of $\overline{CD}$ is between $C$ and $I$.

Here's an image of where the incenter should be located in relation with the angle bisector segment's middle point (that means the section of the angle bisector that's inside of the triangle). What I want to know is why such points cannot be arranged in the order vertex-incenter-middle point.

Because in the standard notation we obtain: $$\frac{CI}{ID}=\frac{BC}{BD}=\frac{a}{\frac{ac}{a+b}}=\frac{a+b}{c}>1$$

• Nice +1 ........ – Aqua Jan 20 '18 at 12:03
• Angle-bisectors meet at $I$. So, by the sine rule, $\frac{CI}a=\frac{\sin CBI}{\sin BIC}=\frac{ID}{BD}$ so $\frac{CI}{ID}=x=\frac{BC}{BD}$. Similarly, $\frac{AC}{AD}=x$. So $\frac{a+b}c=x$ whence your results. – Rosie F Jun 29 '18 at 9:54
• @Rosie F Yes, of course, but firstly I like to try to solve without trigonometry. – Michael Rozenberg Jun 29 '18 at 10:30