∵BB' is the bisector of ∞∠DBC.

∴∠CBB'=∠B'BD (meaning bisector)

∴A∠BA'=∠CBD=2∠B'BD (the opposite vertex angle is equal) ①

Bb' = ab (known)

∴∠BAB'=∠B' (equilateral and equiangular)

∴∠B'BD=∠BAB'+∠B'=2∠BAB' (the outer angle is equal to the sum of two non-adjacent inner angles) ②

Substitute ② into ① to get:

∠ABA'=∠2B'BD=∠4BAB'③

∫AA' is the bisector of ∞∠EAB.

∴∠ EAA' =∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠87

∴∠EAA'+∠A'AB+∠BAB'= 180 degrees (meaning right angle), that is, 2 ∠ AB+∠ Bab' =180 degrees.

That is, ∠ a' ab = (180-∠ Bab') ∠ 2 ④.

Aa' = ab (known)

∴∠AA'B=∠ABA' (equilateral and equiangular)

∴∠A'AB+ABA'+AA'B= 180 degrees (sum of internal angles), that is, 2∠ABA'+∠A'AB= 180 degrees ⑤.

Substitute 3. ④ Enter ⑤ to get:

8 ∠ Bab'+(180-∠ Bab') ∠ 2 =180 degrees.

The solution is ∠BAB' (that is ∠BAC)= 12 degrees.