I am having difficulty solving this problem:
In the given figure (x+y) is an integer greater than 110. What is the smallest possible values of (w+z) (ans is 111)?
Any suggestion on how 111 is the answer?
If you draw a line across the big triangle like the one you drew, but parallel to the base, then we will have $y=w$, so $x+y=x+w$. Since $x+y$ is an integer greater than $110$, the smallest possible value of $x+y$, and hence of $x+w$, is $111$. If the line meets the base somewhere to the right of the base, that is, has negative slope if we think of the base as horizontal, then $w>y$, so $w+x$ will be greater than $111$.
If the line across the triangle has positive slope, then $w$ can be substantially less than $y$, so $w+x$ can be substantially less than $x+y$.
Remark: If we change the problem to asking about $w+z$, then indeed the smallest possible value of $w+z$ is $111$. For the angles complementary to $x$ and $y$ are $180^\circ -x$ and $180^\circ -y$. Thus the sum of the angles of the quadrilateral "below" our line is $(180^\circ-x)+(180^\circ -y)+w+z.$
But the sum of the angles of a convex quadrilateral is $360^\circ$. It follows that $$(180^\circ-x)+(180^\circ -y)+w+z=360^\circ,$$ from which we see that $w+z=x+y$. Since the smallest allowed $x+y$ is $111^\circ$, this is also the smallest allowed value of $w+z$.
Perhaps the simplest way of seeing this is by looking at the angle $B$. Whatever the angles at $A$ (i.e., $x$) and $C$ ($y$) are, you know that $A+B+C$ is $180^\circ$; this means that $x+y+B=180^\circ$. Likewise, since $BDE$ is a triangle, you know that $z+w+B=180^\circ$. But then $z+w = 180^\circ-B = x+y$, and as Andre says in the comments, the smallest integer greater than 110 is 111.