ABCD is a parallelogram, and AXY is a straight line through A meeting BC at X and DC at Y. Prove that BX.DY is constant.

Question

I'm not able to figure out what "constant" here means. I proved it till AB.AD = BX.DY but how is this value constant I don't get it. If we put different values of sides AB and AD we would get different answers. It would be a great help if someone corrects me. Thanks.

Here's the proof :

In ΔABX and ΔYDA

∠XAB = ∠AYD

∠XBA = ∠ADY

Therefore, ΔABX ~ ΔYDA

⇒ AB/DY=BX/AD

Hence, AB.AD = BX.DY