Prove that step by step

In ΔECB ,

∠ECD = ∠EBC + ∠BEC (∵ Exterior Angle Property of a triangle)

⇒ ∠BEC = ∠ECD - ∠EBC   eqn.1

In ΔABC ,

∠ACD = ∠BAC + ∠ABC (∵ Exterior Angle Property of a triangle )

⇒ 2∠ECD = ∠BAC + 2∠EBC

⇒ ∠BAC = 2∠ECD - 2∠EBC = 2(∠ECD - ∠EBC)

( Putting eqn.1 here )

⇒ ∠BAC = 2∠BEC  (∵∠BEC = ∠ECD - ∠EBC)

$12. that is my answer.