AB = AC, and E is the midpoint of BC.

∴AE⊥BC

The dihedral angle A-BC-D is a straight dihedral angle.

∴AE⊥ BCD？

∴AE⊥EF,AE⊥ED

Let BC=2

∠ Barker =∠CDB=90 degrees, ∠DBC=30 degrees.

∴AE=BE=CE= 1

CD= 1

DE= 1

∴AD=√2

∫DF//= CE

∴CDFE is a parallelogram.

∴EF=CD= 1

∴AF=√2

∫DF//BC

∴∠ADF is located between AD and BC.

cosine theorem

cos∠ADF =(2+ 1-2)/(2 *√2 * 1)= 1/2√2 =√2/4

Cosine value of the angle formed by non-planar straight line AD and BC =√2/4.