To find the angle $\angle ADB$ in rhombus ABCD, given $\angle ACB = 40^\circ$, we can use the properties of a rhombus.
Consider triangle $\triangle ABC$. Since ABCD is a rhombus, sides AB and BC are equal (AB = BC). Therefore, $\triangle ABC$ is an isosceles triangle.
In an isosceles triangle, the angles opposite the equal sides are equal. Thus, $\angle BAC = \angle BCA$. Given $\angle ACB = 40^\circ$, we have $\angle BAC = 40^\circ$.
The sum of angles in $\triangle ABC$ is $180^\circ$. So, $\angle ABC = 180^\circ - (\angle BAC + \angle BCA) = 180^\circ - (40^\circ + 40^\circ) = 180^\circ - 80^\circ = 100^\circ$. This is the angle $\angle ABC$.
In a rhombus, the diagonals bisect the angles. The diagonal BD bisects $\angle ABC$. Therefore, $\angle CBD = \frac{\angle ABC}{2} = \frac{100^\circ}{2} = 50^\circ$.
Since opposite sides of a rhombus are parallel, AD || BC. The diagonal BD acts as a transversal line intersecting these parallel lines.
When a transversal intersects parallel lines, alternate interior angles are equal. Therefore, $\angle ADB = \angle CBD$.
From step 4, we know $\angle CBD = 50^\circ$. Thus, $\angle ADB = 50^\circ$.
Using the properties of an isosceles triangle formed by two sides and a diagonal, and the property that opposite sides are parallel, we find that $\angle ADB = 50^\circ$.
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