Solution: ∠ ACB = ∠ b ′+∠ b ′ BC .......... (1), ∫ ab = bb ′, ∴ b ′ = ∠ cab, and substitute into the formula (1).

∠ACB =∠CAB+∠B′BC =∠CAB+( 1/2)∠CBD =∠CAB+( 1/2)(∠CAB+∠ACB)

Multiply both sides by 2 to get 2∠ACB=2∠CAB+∠CAB+∠ACB, ∴ shift term, and merge to get.

∠ACB=3∠CAB, that is ∠ cab = (1/3) ∠ ACB ................................... (2)

∠ABA′=∠CAB+∠ACB =( 1/3)∠ACB+∠ACB =(4/3)∠ACB.........(3)

∠EAA′=∠ACB+∠AA′B =∠ACB+∠ABA′(∫A′A = AB)=∠ACB+(4/3)∠ACB =(7/3)∠ACB

And because A'A shares ∠ EAB ∠ A 'AB = ∠ EAA' = (7/3) ∠ ACB ............. (4).

The sum of the three internal angles of △ aa ′ c is equal to 180, so there is:

∠ACB+∠CAB+∠A′A b+∠AA′B =∠ACB+( 1/3)∠ACB+(7/3)∠ACB+(4/3)∠ACB

=( 1+ 1/3+7/3+4/3)∠ACB = 5∠ACB = 180 ,∴∠acb= 180/5 = 36。