geometryquadrilateral
If 2 quadrilaterals ABCD and PQRS have angles A,B,C,D equal to angles P, Q, R, S respectively and AB=PQ and CD=RS and is AD is not parallel to BC prove that the quadrilaterals are congruent.
I was solving an exercise on the congruence of triangles and I came across this question. Proving the triangles formed by the diagonals to be congruent is certainly not enough but that is all I can think of. What is the condition necessary for two quadrilaterals to be congruent?
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.
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.
Best Answer
We will show that all side lengths are equal
Notation : Line segment $xy=[xy]$ Length of $[xy]=|xy|$
Note that $$ \angle DAB +\angle ABC +\angle ADC +\angle BCD =2\pi $$ so that we have three case :
(1) $\angle DAB +\angle ABC =\pi$ : So $AD\parallel BC$ So it is a contradiction
(2) $ \angle DAB +\angle ABC <\pi $ : There is $X$ s.t. $$ D\in [AX],\ C\in [BX]$$
For $PQRS$ we have $Y$ which is corresponded to $X$
Note that $\triangle XCD$ is congruent to $\triangle YRS$ by SAA-condition (side-angle-anlge - condition)
In further $\triangle XAB$ is congruent to $ \triangle YPQ$ by $SAA$ So $$ |AD|=|AX|-|DX|=|PY|-|SY|=|PS| $$
That is we can show that $|BC|=|QR|$
(3) $ \angle DAB +\angle ABC >\pi $ : This case is completely same to case (2) So we complete the proof