# In Δ ABC ∠B= ∠C and ray AX bisects the exterior angle ∠DAC. If ∠DAX =70°, then ∠ACB =A. 35°B. 90°C. 70°D. 55°

AX bisects DAC

= 140o

By exterior angle theorem,

140o = C + C (Therefore, B = C)

140o = 2C

C = 70o

Therefore, C = ACB = 70o

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