Inequality relation between sines of angles in a triangle. 
Question: Given ${CD}^2=AD.BD$, then prove that, $\sin A.\sin B \le \sin^2{C\over2}$.

I got 2 approaches to the question. Both landing nowhere.

(i) Using sine rule I got that $\sin A.\sin B=\sin \alpha.\sin \beta$ but can't proceed.
(ii) On extending $CD$ further to $D'$ such that $CD=DD'={CD'\over2}$ which makes $ABCD'$ a cyclic quadrilateral again ending where the first step ends.

Help me please.

Thanks for any hints or solution. Hope that question is decently put.
 A: Taking from where you left off,and using the well-known fact:  $xy \le \dfrac{(x+y)^2}{4}$, we have : $ \sin A\cdot \sin B = \sin \alpha \cdot \sin \beta  \le  \dfrac{ (\sin  \alpha + \sin \beta )^2}{4} = \dfrac{\left(2\sin(\frac{\alpha+\beta}{2})\cdot \cos(\frac{\alpha - \beta}{2})\right)^2}{4} = \sin^2(\frac{\alpha+\beta}{2})\cdot \cos^2(\frac{\alpha-\beta}{2}) \le \sin^2 (\frac{\alpha+\beta}{2}) = \sin^2 (C/2)$  because  $\cos^2(\frac{\alpha  -  \beta}{2}) \le  1$. This establishes the claim.
A: DeepSea's answer is nice, but it seems that you don't get it.
So, let me add more details.
You've already got 
$$\sin A\sin B=\sin\alpha\sin\beta\tag1$$
Since
$$\frac{(x+y)^2}{4}-xy=\frac{x^2+2xy+y^2-4xy}{4}=\frac{(x-y)^2}{4}\ge 0$$
holds for any $x,y\in\mathbb R$, we see that
$$xy\le\frac{(x+y)^2}{4}$$
holds for any $x,y\in\mathbb R$.
So, we have
$$\sin\alpha\sin\beta\le \frac{(\sin\alpha+\sin\beta)^2}{4}\tag2$$
Using the following sum-to-product identity
$$\sin\alpha+\sin\beta=2\sin\bigg(\frac{\alpha+\beta}{2}\bigg)\cos\bigg(\frac{\alpha-\beta}{2}\bigg)$$
one gets
$$\begin{align}\frac{(\sin\alpha+\sin\beta)^2}{4}&
=\frac{1}{4}\left(2\sin\bigg(\frac{\alpha+\beta}{2}\bigg)\cos\bigg(\frac{\alpha-\beta}{2}\bigg)\right)^2
\\\\&=\sin^2\bigg(\frac{\alpha+\beta}{2}\bigg)\cos^2\bigg(\frac{\alpha-\beta}{2}\bigg)\tag3\end{align}$$
For any $\theta\in\mathbb R$, we have $|\cos\theta|\le 1$, so $\cos^2\theta\le 1$.
So, we have
$$\cos^2\bigg(\frac{\alpha-\beta}{2}\bigg)\le 1$$
Multiplying the both sides by $\sin^2(\frac{\alpha+\beta}{2})\ (\gt 0)$ gives
$$\sin^2\bigg(\frac{\alpha+\beta}{2}\bigg)\cos^2\bigg(\frac{\alpha-\beta}{2}\bigg)\le \sin^2\bigg(\frac{\alpha+\beta}{2}\bigg)=\sin^2\frac{C}{2}\tag4$$
It follows from $(1)(2)(3)(4)$ that
$$\sin A\sin B\le\sin^2\frac C2$$
