geometry
A quadrilateral with one diagonal that bisects opposite angles and
another diagonal that does not bisect opposite angles may or may
not be a kite.
We know that a kite is a quadrilateral if it has two pair of equal and adjacent sides, of which the diagonals intersects at right angles, but how could we conclusively prove the validity of this statement?
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.
As stated the problem can have many solutions.
For example:
Let the quadrilateral $ABCD$ such that $AD=DB=BC=x$ and $AC=\sqrt{3}x$. See figure 1.
Figure 1
The angle $b$ depends on $a$ as the expression:
$$b = \frac{a}{2}- \frac {\pi}{2}+ \arccos(- \frac{1}{2 \sin {\frac{a}{2}}}+ \sin {\frac{a}{2}})$$
So we have many solutions, for example: $a=\frac{\pi}{3}$ and $b=\frac{\pi}{3}$ or $a=\frac{\pi}{4}$ and $b=\frac{\pi}{2}$ as shown in figure 2.
Figure 2
Best Answer
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.