Often while solving mensuration problems, I encounter situations where I need to establish the congruence of two quadrilaterals, mostly having been got all the four sides (and of course they are equal for both of them). This is where I get entangled.
[Assuming I have the sides in a particular order.] I have proved that when the quadrilaterals are a trapezium (but not a parallelogram), they are congruent. However, when it is not so, and I have other information such as some angles and diagonals. For example, visualizing a construction, I can see that having one diagonal gives us two cases, and thus not unique. But I am not able to prove that having two diagonals also gives us a unique case. With an angle or two opposite angles there are again two cases but I am not able to see anything else. I know that assuming that it a convex quadrilateral, simplifies the problem considerably, but it is not necessary.
So, my question is: Can anyone give me a characterization of when two quadrilaterals are congruent, knowing the sides and that they are equal? And how much does assuming that the quad is convex help us?
I have another doubt, let us suppose that the area of a type of quadrilateral is $f(a,b,c,d,e)$, where the letters are some elements of the quadrilateral. Then knowing those particular elements, make the quadrilateral unique? If this is true, this simplifies some problems immensely, such as the uniqueness of cyclic quadrilaterals given sides in order, which I have not been able to do otherwise.
Best Answer
Given two quadrilaterals $ABCD$ and $EFGH$, with side lengths equal, (i.e. $d(A,B)=d(E,F)$, etc.) then $ABCD \cong EFGH$ if and only if the diagonals have equal length: $d(A,C)=d(E,G)$ and $d(B,D)=d(F,H)$, where $d$ is the distance between the points.
The forward implication is obvious. To see the converse, we first notice that the diagonal's length being equal gives us a few congruent triangles. Namely, $\triangle ABC \cong \triangle EFG$, $\triangle ACD \cong \triangle EGH$ from $d(A,C)=d(E,G)$, by the good old SSS property.
Using only the one diagonals being equal, we would have two possible quadrilaterals from these facts. They come from reflecting one of the triangles over the diagonal.
Next, we see that we are given the distance between $D$ and $B$. Therefore, we are forced in which of these two options must be our congruent quadrilateral. The only way that both quadrilaterals could be acceptable is if $B$ is the proper distance from $D$ in both. But this will force $B$ onto the diagonal between $A$ and $C$, which contradicts the definition of a polygon (giving you a triangle instead), and finishes the proof.
I hope this was helpful. I believe a slight variation of this will work if you know the length of one diagonal, say $d(A,C)$ and the angle $\angle BCD$ or $\angle BAD$. It would seem reasonable that the same will hold if you know area, the four sides and a diagonal also. The fact that you have two possibilities right away from one triangle gives you most of the information. Also, it is pretty clear that the reflection over one of the diagonals will always produce one convex quadrilateral and one not.