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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.

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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.

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    $\begingroup$ Note that with the additional constraints that $|AB|=|BC|$ and $|CD| = |DA|$, the quadrilateral is forced to be a kite, since now we have side-angle-side congruency of the two triangles formed by splitting along the diagonal $BD$. A nice additional puzzle: what if we're only given that $|AB|=|BC|$? $\endgroup$ – Steven Stadnicki Mar 14 '14 at 20:48
  • $\begingroup$ If $ΑΒ = ΒС$, then by cosine rule $CD = DA$, cause $\angle A = \angle C$, isn't it&? $\endgroup$ – user135508 Mar 14 '14 at 20:53
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    $\begingroup$ Basically, yep - in this case we have side-side-angle congruency, since we know that the (common) side BD of the triangles ABD and CBD is the same, we're given that AB/CB are the same and we have one common angle. $\endgroup$ – Steven Stadnicki Mar 14 '14 at 21:46
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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.

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