Help me prove congruency 
i) 
$AX = BC$ (given)
$AD = CY$ (given since $AX = CY = BC$ and $BC = AD$ in a parallelogram)
$\widehat{DAX} =  \widehat{YCB}$ (equal opp. angles in a parallelogram)
Therefore $ADX$ and $CBY$ are congruent
Now this is my solution to this question, is it mathematically valid?

The line "$\angle DAX=\angle YCB$ since equal opp angles in parallelogram" needs some work.
Either you are referring to some rule I have not heard of, or else you are misapplying the one I have heard of, which would say that $\angle ACX=\angle YBC$. I don't think it's the first one because a property of parallelograms would not refer to these extra points $X,Y$ which are not a common feature of parallelograms.
To get you going again on a solution, work to show $AX\parallel YC$. here are two hints:
(I'm also going to assume that the diagram is meant to further convey the hypothesis that $X$ does not lie directly on $D$.)
Hint 1: Look at the two isosceles triangles.
Hint 2: Look at $\angle BYC$ and $\angle AXD$.
