If convex quadrilateral ABCD has congruent angles A and C and the other pair of angles B and D are not congruent, is ABCD necessarily a kite (two pairs of consecutive congruent sides but opposite sides are not congruent)? I've tried proof by contradiction, that B and C are congruent, but I can't seem to find any contradiction coming from that. FWIW I've constructed a figure like this in GeoGebra, and it's a kite, as I suspected, but I'm having trouble proving it. Drawing diagonals doesn't lead me to anything fruitful. I feel that this should be fairly straightforward, but I can't crack this nut.
No. Take a circle with diameter BD, and let A, C be any points on it. So A and C will be right angles and all right angles are congruent, but this isn't true in general that ABCD would be a kite.
Given $A$ and $C$ join them and construct the perpendicular bisector $L$ of $AC$. Let $O_B$ be a point on $L$ distance $d$ from $AC$ so that $O_BA=O_BC=r$, and on the other side of $AC$ construct $O_D$ on $L$ also distance $d$ from $AC$ and distance $r$ from $A$ and $C$.
On the two congruent (by symmetry) arcs joining $A$ and $C$ with centres $O_B$ and $O_D$ choose arbitrary points $B$ and $D$. It will be found that the angle $ABC$ is the same as the angle $ADC$ whichever points are chosen (angles in congruent arcs are equal).
Alternatively cut a piece of paper to give you two corners with the same angle - right angles are especially easy - and give it a try.
Alternatively consider how many rectangles are kites (I know all the angles are the same) - and then flex the rectangle a little so two opposite angles change.