<Qua-2024-741>, hint

Let AC and BD meet at Q, AD and BC meet at P, Line PQ cuts CD at M, AB at N .

Let CD=x AB=y. Apply Menelaus theorem to triangle ACD with traverse PMQ.

(CM/MD)DP/PA)(AQ/QC)=1 ...>(CM/MD)(x/y)(y/x)=1, Hence CM=MD/

Similarly AN=NB.

Triangles PDC and  PAB are similar, and triangles QCD and QAB are similar