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?
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$\begingroup$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.
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