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.
[Math] Is a quadrilateral with one pair of opposite angles congruent and the other pair noncongruent necessarily a kite
euclidean-geometrygeometryproof-writingquadrilateral
Related Solutions
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.
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
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.