Corresponding angles of $\square ABCD$ and $\square PQRS$ are equal, $AB=PQ$, $CD=SR$, $AD\not\parallel BC$; prove the quadrilaterals congruent

Question

If two quadrilaterals $ABCD$ and $PQRS$ have angles $A$, $B$, $C$, $D$ equal to angles $P$, $Q$, $R$, $S$ respectively and $AB=PQ$, $DC=SR$, and if $AD$ is not parallel to $BC$, prove that the quadrilaterals are congruent.

What I have done - Given all angles are equal, we have to prove that $BC=QR$ and $AD=PS$. I have tried using congruency of triangles as follows: $AB=PQ$, and $\angle A$ is equal to $\angle P$. However, I cannot find the third equation to complete the proof.

Can someone please help?

Answer 1

hint: first you need to prove PS is not parallel to QR.

then make AE//BC, PT//QR, prove AE=PT, AC=PR, then ....

Answer 2

Extend DA and CB to meet at E. Similarly, extend SP and RQ to meet at T. In ∆EDC and ∆TSR, angle EDC = angle TSR, and angle ECD = angle TRS. Also, DC = SR. Thus, both the triangles are congruent, by ASA axiom. Apply same congruency in ∆EAB and ∆TPQ. Now, it will be obvious that both the quadrilaterals should be congruent