Since I can't draw, I will use coordinates, and you can do the drawing.
The quadrilateral clearly can be a kite. For completeness, we show this. Let the vertices of our quadrilateral, in counterclockwise order, be $A(1,0)$, $B(0,2)$, $C(-1,0)$, and $D(0,-1)$. This is a kite, and the diagonal $BD$ bisects a pair of opposite angles, and the diagonal $AC$ doesn't.
Now let's produce a suitable non-kite $ABCD$. What is a kite? Does it have to be convex? If it does, here is an example of a non-kite with the desired properties. Let the vertices be $A(1,0)$, $B(0,2)$, $C(-1,0)$, and $D(0,1)$. Note that this is non-convex, the part $CDA$ sticks in, not out.
If you are limiting attention to convex quadrilaterals, then, as was pointed out by Henry, there are no non-kites which have the property that one diagonal bisects a pair of opposite angles. For the diagonal that bisects a pair of opposite angles divided the quadrilateral into two triangle, which can be shown to be congruent (they have a common side, and all corresponding angles match). Thus sides match in pairs. If the quadrilateral is convex, this forces it to be a kite.
In a possible attempt to explain a), let us focus solely on a single angle, say angle $A$. Similarly, draw tangent lines extending from the two adjacent sides, namely $AB$ and $AD$. Assuming $A\neq180$ (which we can, because it would cause $ABCD$ to be a triangle), $AB$ and $AD$ are not parallel.
This means that they meet at $A$ and continue, getting further apart as they go. If $A \lt 180$, meaning $ABCD$ is convex, $AB$ and $AD$ continue away from the shape, not intersecting any sides.
However, if $A \gt 180$, $AB$ and $AD$ enter the interior or $ABCD$ after intersecting at $A$. As the lines are infinite and the quadrilateral is not, the lines must at some point leave the shape. As two lines can only meet at a single point, and will not intersect themselves, they must leave the shape through one of the other two sides (Note Pasch's Theorem).
As both $AB$ and $AD$ are equally dependent on the angle of $A$, it is not possible for only one of the two lines to split one of the other sides.
Best Answer
This is a complete proof, and it might get a bit repetitive because I'm using the Sine Law again and again. I'm trying to find something more elegant, but this is the only thing thing I could come up with right now.( I've used a different diagram because my internet wasn't working while I was working this out. Just skip over to the ending note if you just want to see how the angles are equal.)
$DEGF$ is the quadrilateral and $A$ and $B$ are the foci.
Using sine law in $\Delta DBE$ you get $\dfrac{DE}{BE}=\dfrac{\sin\angle DBE}{\sin\angle BDE}$ and in $\Delta GBE$ you get $\dfrac{GE}{BE}=\dfrac{\sin\angle GBE}{\sin\angle BGE}$. Dividing the two you'll obtain: $\dfrac{DE}{GE}=\dfrac{\sin\angle DBE}{\sin\angle BDE}\cdot\dfrac{\sin\angle BGE}{\sin\angle GBE}\tag{i}$
Using sine law in $\Delta BDF$ you get $\dfrac{FD}{FB}=\dfrac{\sin\angle DBF}{\sin\angle BDF}$ and in $\Delta BGF$ you get $\dfrac{FG}{FB}=\dfrac{\sin\angle GBF}{\sin\angle BGF}$. Dividing the two you'll obtain: $\dfrac{FD}{FG}=\dfrac{\sin\angle DBF}{\sin\angle BDF}\cdot\dfrac{\sin\angle BGF}{\sin\angle GBF}\tag{ii}$
Dividing (i) and (ii):$\dfrac{DE\cdot FG}{GE\cdot FD}=\dfrac{\sin\angle BGE}{\sin\angle BDE}\cdot\dfrac{\sin\angle BDF}{\sin\angle BGF}\tag{iii}$
Using sine law in $\Delta DFI$ you get $\dfrac{DI}{FI}=\dfrac{\sin\angle DFE}{\sin\angle FDG}$ and in $\Delta GFI$ you get $\dfrac{GI}{FI}=\dfrac{\sin\angle GFE}{\sin\angle FGD}$. Dividing the two you'll obtain: $\dfrac{DI}{GI}=\dfrac{\sin\angle DFE}{\sin\angle FDG}\cdot\dfrac{\sin\angle FGD}{\sin\angle GFE}\tag{iv}$
Using sine law in $\Delta DEI$ you get $\dfrac{DI}{EI}=\dfrac{\sin\angle DEF}{\sin\angle EDG}$ and in $\Delta GEI$ you get $\dfrac{GI}{EI}=\dfrac{\sin\angle GEF}{\sin\angle EGD}$. Dividing the two you'll obtain: $\dfrac{DI}{GI}=\dfrac{\sin\angle DEF}{\sin\angle EDG}\cdot\dfrac{\sin\angle EGD}{\sin\angle GEF}\tag{v}$
Combining (iv) and (v): $\dfrac{DI}{GI}=\dfrac{\sin\angle DFE}{\sin\angle FDG}\cdot\dfrac{\sin\angle FGD}{\sin\angle GFE}=\dfrac{\sin\angle DEF}{\sin\angle EDG}\cdot\dfrac{\sin\angle EGD}{\sin\angle GEF}\tag{vi}$
Doing the same thing with $DB$ and $GB$ you will get $$\dfrac{DB}{GB}=\dfrac{\sin\angle DFE}{\sin\angle GFE}\dfrac{\sin\angle BGF}{\sin\angle BDF}=\dfrac{\sin\angle DEF}{\sin\angle GEF}\dfrac{\sin\angle BGE}{\sin\angle BDE}\tag{vii}$$
Dividing (vi) and (vii) you get:$$\dfrac{\sin\angle FGD}{\sin\angle FDG}\cdot\dfrac{\sin\angle BDF}{\sin\angle BGF}=\dfrac{\sin\angle DGE}{\sin\angle GDE}\cdot\dfrac{\sin\angle BDE}{\sin\angle BGE}\tag{viii}$$
Now due to the Property 1 on this link, $\angle FGD=\angle BGE, \angle GDE=\angle BDF,\angle DGE=\angle BGF$ and $\angle BDE=\angle FDG$. Making the necessary substitutions, relation (viii) becomes:$$\sin^2\angle BGE \cdot\sin^2\angle BDF=\sin^2\angle BDE \cdot\sin^2\angle BGF$$ $$\sin\angle BGE \cdot\sin\angle BDF=\sin\angle BDE \cdot\sin\angle BGF\tag{ix}$$
Combining (iii) and (ix): $DE\cdot FG=GE\cdot FD$
NOTE: Making use of the relations (vi),(vii) and (ix) it is easy to see that $\dfrac{DI}{GI}=\dfrac{DB}{GB}$. By the Angle Bisector Theorem, this means that $\angle DBF = \angle GBF$ and similarly $\angle FAG = \angle EAG$.