BC=2CD, take the midpoint m of BC and connect FM.

It is conceivable that the triangle FBC is an isosceles triangle.

Triangle ABD is an isosceles triangle.

Angle BFC= Angle BAD= Angle BAC+ Angle CAD= Angle BAC+ Angle ABF

Angle ABF= Angle DAC= Angle DBC

Angle ABD= angle FBC

Angle difference = angle BFC

Triangle ABD is similar to triangle FBC.

The triangle FBC is an isosceles triangle with FB=FC.

Take the midpoint m of BC and connect FM.

Then angle MFC= angle BFC/2= angle DFC, angle BCA= angle DCA, and cf = cf

The triangle FMC is equal to FDC.

Angle FDC= angle FMC=90 degrees, FD is perpendicular to CD.

BC=2CD