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I want to construct a quadrilateral which is not a parallelogram, in which a pair of opposite angles and a pair of opposite sides are equal. I tried drawing one, but I am not able to. Please help. (And I dont think I will be able to post what I have tried or drawn. Hope the statement of the post is clear).

Umm, I understand that it might look like it is the duplicate of another similar question( I saw that post actually!). But I do not think it explained how to construct one. I know such a quadrilateral exists, that is not my real question. I want to know how it is constructed. I hope I made it clear.

Also, I wanted a short, simplified and clear solution for this(I can elaborate that).

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2 Answers

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Take $\Delta ABK$ such that $\measuredangle ABK>90^{\circ}$ and let $D\in AK$ such that $BD=BK$.

Now, let $\Delta BDC\cong\Delta KBA$ such that $K$ and $C$ are placed in the same side respect to the line $DB$.

Thus, $ABCD$ is a needed quadrilateral because $AB=DC$ and $\measuredangle A=\measuredangle C$.

Easy to see that $ABCD$ is not parallelogram.

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You are not succeeding in finding such a quadrilateral because none exists.
If $ABCD$ is such a quadrilateral with $|AD| = |BC|$ and $\beta = \delta$ then the triangles $ABC$ and $CDA$ are congruent. It follows that $\sphericalangle CAB = \sphericalangle DCA$, hence $AB$ is parallel to $DC$ from this you can easily deduce that the quadrilateral is a parallelogram.

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