In ΔABC, AD is the median and AD = (1/2)BC. If ∠ACD = 40°, then what is the value (in degrees) of ∠DAB?

AD = (1/2)BC and AD is the median which bisects BC at D, => AD = BD = CD

=> $$\triangle$$ ADC and $$\triangle$$ ABD are isosceles triangles.

Let $$\angle$$ ACD = $$\angle$$ CAD = $$\theta=40^\circ$$ -------------(i)

=> $$\angle$$ ADC = $$(180$$°$$ - 2 \theta)$$

Now, $$\angle$$ ADB and $$\angle$$ ADC are supplementary

=> $$\angle$$ ADB + $$\angle$$ ADC = 180°

=> $$\angle$$ ADB = $$180$$°$$ - (180$$°$$ - 2 \theta) = 2 \theta$$

Also, $$\angle$$ ABD = $$\angle$$ BAD  = $$x$$

In $$\triangle$$ ABD, => $$\angle$$ ABD + $$\angle$$ BAD + $$\angle$$ ADB = 180°

=> $$x + x + 2 \theta = 180$$°

=> $$2(x + \theta) = 180$$°

=> $$x + 40^\circ = \frac{180}{2}=90^\circ$$     [Using equation (i)]

=> $$x=90-40=50^\circ$$

=> Ans - (C)