Draw a parallel line from d parallel to BC and construct ∠DCE=20? Parallel lines intersect at point e to connect AE.

Let's make a point E about the symmetry point E' of the edge AC, and the intersection of connecting EE', AE', CE', EE' and AC is F.

∫ Germany∨ BC

∴∠EDC=∠BCD=20？

∫∠DCE = 20？

∴∠EDC=∠DCE, and then ED=EC.

∫ de ∨BC, ∠DBC=40? And < ECB = < ECD+< BCD = 20? +20? =40?

∴ trapezoid DBCE is isosceles trapezoid, then BD=CE.

∠∠ABC =∠ABD+∠DBC = 70？

∠ACB=∠ACD+∠DCB=70？

∴AB=AC

∫∠ACD = 50？ ，∠DCE=20？

∴∠ACE=∠ACD-∠DCE=30？

In △ABD and △ACE:

AB=AC

∠ABD=∠ACE=30？

BD=CE

∴△ABD≌△ACE

∴AD=AE,∠BAD=∠CAE

It can be easily proved by symmetry: △ AFE △ AFE' (SAS)

∴AE=AE', and then AD=AE'

∫∠ACE = 30？ Namely: ∠FCE=30?

∴:2ef = ce in Rt△EFC.

EE ' = 2EF

∴CE=EE'

From CE=DE: EE'=DE

In △ADE and △AE'E:

' AD=AE '，AE=AE，DE=EE '

∴△ADE≌△AE'E (SSS)

∴∠AED=∠AEE'

∫In△CED:∠DCE = 20？ ，CE=DE

∴∠CED= 180？ -2∠DCE= 140？

∴∠AED+∠AEE'=2∠AED=360？ -∠CED-∠CEF

=360? - 140? -60? = 160?

∴△ade:∠DAE = 180 in isosceles triangle? -2∠AED= 180？ - 160? =20?

∴∠BAD+∠CAE= 180？ -2∠ABC-∠DAE

So 2∠BAD= 180? -2×70? -20?

∴∠BAD= 10？