In Δ ABC, the median AD is 8 cm and CB is 16 cm, measure of angle CAB is_____ .

BC = 16 cm and AD is the median which bisects BC at D, => BD = DC = 8 cm

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

Let $$\angle$$ ACD = $$\angle$$ CAD = $$\theta$$

=> $$\angle$$ 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°

=> $$x + x + 2 \theta = 180$$°

=> $$2(x + \theta) = 180$$°

=> $$(x + \theta) = \frac{180}{2}$$

=> $$\angle$$ CAB = 90°