Question: As shown in the figure, in triangle ABC, DE is on BC, BD=AB, CE=AC, ∠DAE= 1/3∠BCA, ∠BAC.

Answer: BD=AB, CE=AC.

So ∠ bad = ∠ BAE+∠ EAD = ∠ ade ( 1)

So ∠CAE=∠CAD+∠EAD=∠AED②.

At delta △AED.

∠AED+∠ADE+∠DAE= 180 ③

∠DAE= 1/3∠BAC④

According to four formulas

Available BAC+2 EAD = 5 EAD = 180.

So ∠ BAC = 3 ∠ EAD = 108.