Q6 of 69 Page 9

In Δ ABC ∠B= ∠C and ray AX bisects the exterior angle ∠DAC. If ∠DAX =70°, then ∠ACB =

AX bisects ∠DAC

∠CAD = 2 * ∠DAC


∠CAD = 2 * 70o


= 140o


By exterior angle theorem,


∠CAD = ∠B + ∠C


140o = ∠C + ∠C (Therefore, ∠B = ∠C)


140o = 2∠C


∠C = 70o


Therefore, ∠C = ∠ACB = 70o

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