46 In $$\triangle$$ ABC, DE $$\parallel$$ BC in such a way that A-D-B and A-E-C. If m $$\angle ACB = 40^{\circ}$$, then m $$\angle$$ DAE + m $$\angle$$ ADE = __________

A

$$240^{\circ}$$

B

$$120^{\circ}$$

C

$$140^{\circ}$$

D

$$230^{\circ}$$