Relation between the GM of two sides of a triangle and the bisector of angle between them I am trying to solve the following multiple choice problem :
Let the bisector of the angle $C$ of a triangle $ABC$ intersect the side $AB$ at a point $D$. Then the geometric mean of $CA$ and $CB$ 


*

*is less than $CD$

*is equal to $CD$

*is greater than $CD$

*does not always satisfy any one of the foregoing properties.


I have tried that following steps :

Since $CD$ bisects $\angle C$, so $\angle ACD=\angle DCB$ and $\dfrac{CA}{CB}=\dfrac{AD}{DB}$.
Also, $ext\angle ADC=\angle DCB+\angle DBC$, so that $\angle ADC>\angle DCB=\angle ACD$. This shows that $CA>AD$. Similarly, $CB>DB$. Now, by triangle inequality, $CA+AD>CD$ and $CB+DB>CD$. Thus it follows that $2CA>CD$ and $2CB>CD$ and this gives me $\sqrt{CA\cdot CB}>\dfrac{CD}{2}$. But the answer as given in my textbook is $\sqrt{CA\cdot CB}>CD$ i.e. the option $3$. How do I proceed to solve this problem? 
 A: 
Lemma. $CD<\sqrt{CA\cdot CB}$    
Proof.
$\triangle ADC\sim\triangle ABE$ $\Longrightarrow$ $BE=\dfrac{CB+CA}{CA}\cdot CD<CB+CE=2CB$    
$\therefore$ $CD<\dfrac{2CA\cdot CB}{CA+CB}\leq\dfrac{2CA\cdot CB}{2\sqrt{CA\cdot CB}}=\sqrt{CA\cdot CB}$
A: Thanks to alex.jordan for the valuable comment ! I think I have an answer now:

As suggested, I have made the necessary construction. I have produced $AC$ to $AB'$ such that $CB'=CB$ and then produced $DC$ to $DD'$ such that $DC=CD'$. Join $B'$ and $D'$. Then clearly $\Delta BCD\cong \Delta B'CD'$, so that $\angle D'B'C=\angle DBC$. Then I have drawn the circumcircle of $\Delta ADB'$. Let the line joining $D$ and $D'$ intersect this circle at $D_1$. I need to show that $D'$ lies between $D$ and $D_1$. To the contrary let us suppose that $D'$ lies at the position $D_2$ which is outside the circle. Then by construction, $\Delta B'CD_2\cong \Delta BCD$, so that $\angle D_2B'C=\angle D_2B'A=\angle DBC$. But then $\angle D_2B'A>D_1B'A=\angle D_1DA$ which will imply that $\angle DBC>\angle ADC$ but this is not true because $\angle ADC=\angle BDC+\angle DCB$ so that $\angle ADC>\angle DBC$. Thus, $D'$ cannot be outside the circle. similarly we can show that $D'$ cannot lie on the circle too. Hence $D'$ should lie inside the circle. Thus, $CD'<CD_1$. Hence, $CA\cdot CB=CD\cdot CD_1>CD.CD'=CD^2$. Thus, geometric mean of $CA$ and $CB$ is always greater than $CD$.
But, is there any other easier way to solve this problem? 
