Phase difference for a central maxima in diffraction 
For a central maximum in diffraction, contributions from all the parts of slits like point C and D are supposed to be in phase. I am not getting why contributions from all parts of slits should in phase. I think phase difference will depend on path difference and looks like there will be path difference between different parts of slit like CD and AB will have path difference as I show in picture. 
 A: You are correct that there will be a path difference comparing AB and CB.  When we assume that all the waves arrive at B in phase we are using the far field approximation.  Diffraction analyzed this way is referred to as Fraunhofer diffraction. This does not hold if point B is near the aperture, in that case you need to use the more complicated Fresnel diffraction equation.
A: You are right that the contributions from different parts of the slit will be rigorously out of phase. However, at long distance $L$ from the slit, you can show that the path length different between two points $A$ and $C$ can be written $\delta_{A,C} = n \overrightarrow{AC} \cdot \overrightarrow{u}_{AC, B}$, where $\overrightarrow{u}_{AC, B}$ is the unit vector from the middle of $[AC]$ pointing towards $B$. But at very large distances, if you look around the center of the screen, $\overrightarrow{u}$ is basically horizontal, while $\overrightarrow{AC}$ is vertical, meaning that their dot product is $0$. 
The next contribution is second order, which is neglected here (true if $a^2/L << \lambda$, where $a$ is the size of the slit). For instance, for a quite large $1\,\mathrm{mm}$ slit, and $\lambda$ in the visible range, the LHS will be smaller than the RHS for $L >> 1\,\mathrm{mm}$ (this is the criteria for being in the "far-field" approximation).
As you can see in the image below, constructive interferences at the center of the x axis do not occur for $\frac{L \lambda}{a^2} \lessapprox 1$ (near-field), but do occur for larger distances.

