$$ABCD$$ is a quadrilateral. $$AB$$ is parallel to $$CD$$ and $$AB>CD$$. If $$AD=AB=BC$$ and $$\angle ADC=140^\circ$$, then the measure of $$\angle CAB$$ is ______ degrees.

Since $$AB\parallel CD$$, the consecutive interior angles give $$\angle DAB=180^\circ-140^\circ=40^\circ$$. Also, $$AD=BC$$ makes the trapezium isosceles, so $$\angle ABC=40^\circ$$. In triangle $$ABC$$, $$AB=BC$$, so if $$\angle CAB=\angle ACB=x$$, then $$2x+40^\circ=180^\circ$$ and $$x=70^\circ$$.