# Construct or prove existence of a certain quadrilateral

• It is convex.
• It has exactly one pair of congruent opposite sides.
• It has exactly one pair of congruent opposite angles.
• It is not a parallelogram.

Does such a quadrilateral exist? Is it possible to construct such a quadrilateral with compass and straightedge? If the construction is not possible, why not?

• Note -- the "no parallelogram" restriction is redundant. Also, you the word "exactly" in the opposite angle condition isn't needed. Commented Jan 19, 2017 at 16:07

Yes, such quadrilaterals exist -- even constructible ones.

For example, the quadrilateral $ABCD$ with \begin{align*} \angle DAB &= 60^\circ\\[6pt] \angle ABD &= 45^\circ\\[6pt] \angle ADB &= 75^\circ\\[6pt] \angle BCD &= 60^\circ\\[6pt] \angle CBD &= 105^\circ\\[6pt] \angle CDB &= 15^\circ\\ \end{align*}

satisfies the required conditions.

Let ABC be an isosceles right triangle with AC = BC. On base AB make BD = AC and join CD. Thus angle BCD = angle BDC = 67.5 degrees. On CD, congruent to triangle CDB, construct triangle CDB' in a plane perpendicular to that of triangle ACD, forming quadrilateral ACB'D. Since angle B'CA is less than 90 degrees (it's less than the angle AC makes with a perpendicular to the plane of ACD at A), while angle B'DA is greater than 90 degrees (it's greater than the angle AD makes with a perpendicular to ACD at D), these angles are not equal. But angle CAD = opposite angle CB'D = 45 degrees. And AC = opposite side B'D. But AC is not parallel to B'D, since they are not co-planar, and for the same reason CB' is not parallel to AD. Therefore, a quadrilateral ACB'D has been constructed (with straightedge and compass) in which only one pair of opposite sides and angles are equal, and which is not a parallelogram.