<Qua-2024-734>, hint

Let AB=a, BC=b, <ABD=x, <DBC=y

BC=b=2 a cosy, AB=a=2b cos(x+y) Hnce  cosy cos(x+y)=1/4

BE=2 AB cos(x+y)=2a cos(x+y),  BD=a =2 BE cosy Triangle BED is isosceles(BE=BD)

Hence a= 2 2a cosy  cos(x+y). cosycos(x+y)=1/4