# prove that the quadrilaterals are congruent

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?

• See if this link is helpful : mcs.uvawise.edu/msh3e/resources/geometryBook/… – астон вілла олоф мэллбэрг Oct 8 '16 at 3:38
• @астонвіллаолофмэллбэрг This link directly states some congruence conditions but I need a proof based on the congruence of triangles. – oshhh Oct 8 '16 at 3:47
• Aren't the proofs given in the latter part of the document? – астон вілла олоф мэллбэрг Oct 8 '16 at 3:48
• @астонвіллаолофмэллбэрг I didn't see that. Thanks! But in this question is there any other simpler way to solve this because here I can't simply apply this congruence, I would have to prove it first... – oshhh Oct 8 '16 at 3:53
• No, I would think it is difficult to apply it. It is a difficult problem, because of the assmuption on the non-parallellity of $AB$ and $CD$, which doesn't seemingly have anything to do with congruence. – астон вілла олоф мэллбэрг Oct 8 '16 at 3:59

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