Answer:
proof below
Explanation:
∠A ≅ ∠B
as ACB is a triangle
if ∠A ≅ ∠B then AC = BC
and CD ⊥ AB
so ∠CDA = ∠CDB = 90
now in ΔADC ---> ∠A + ∠CDA + ∠ACD = 180
lets assume ∠A = x = ∠B
x + 90 + ∠ACD = 180
∠ACD = 180 - 90 - x
∠ACD = 90 - x
now in ΔCDB ---> ∠B + ∠CDB + ∠BCD = 180
x + 90 + ∠BCD = 180
∠BCD = 180 - 90 - x
∠BCD = 90 - x
As ∠ACD = ∠BCD
CD bisects ∠ACB