Let $$ABCD$$ be a quadrilateral with $$\angle ADC = 70^\circ$$, $$\angle ACD = 70^\circ$$, $$\angle ACB = 10^\circ$$ and $$\angle BAD = 110^\circ$$. The measure of $$\angle CAB$$ (in degrees) is:

Join the diagonal $$AC$$ so that the quadrilateral $$ABCD$$ is split into the two triangles $$\triangle ABC$$ and $$\triangle ACD$$.

Step 1: Work inside $$\triangle ACD$$
The two given angles at $$D$$ and $$C$$ are each $$70^\circ$$. Hence the third angle is

$$\angle CAD \;=\;180^\circ - (70^\circ+70^\circ)=40^\circ.$$

Step 2: Relate the angle at vertex $$C$$ of the quadrilateral
The interior angle of the quadrilateral at $$C$$, denoted $$\angle BCD$$, is the sum of the two angles that meet at $$C$$ along the diagonal:
$$\angle BCD=\angle BCA+\angle ACD = 10^\circ+70^\circ = 80^\circ.$$

Step 3: Use the angle-sum of the quadrilateral
For any quadrilateral,
$$\angle BAD+\angle ABC+\angle BCD+\angle CDA = 360^\circ.$$
Insert the known values: $$\angle BAD=110^\circ,\; \angle BCD=80^\circ,\; \angle CDA=\angle ADC=70^\circ.$$
$$110^\circ+\angle ABC+80^\circ+70^\circ = 360^\circ \\ \Longrightarrow \angle ABC = 360^\circ-260^\circ = 100^\circ.$$

Step 4: Angle-sum in $$\triangle ABC$$
Let $$\angle CAB = x$$. In $$\triangle ABC$$:
$$x + \angle ABC + \angle ACB = 180^\circ,$$
$$x + 100^\circ + 10^\circ = 180^\circ,$$
$$x = 180^\circ-110^\circ = 70^\circ.$$

Hence $$\angle CAB = 70^\circ.$

Final Answer: $$70^\circ$$