To find the angle $\angle ADB$ in rhombus ABCD, given $\angle ACB = 40^\circ$, we can use the properties of a rhombus.
Rhombus Angle Properties Utilized
- All sides of a rhombus are equal in length (AB = BC = CD = DA).
- Opposite angles are equal ($\angle DAB = \angle BCD$, $\angle ABC = \angle ADC$).
- Adjacent angles sum to $180^\circ$ ($\angle DAB + \angle ABC = 180^\circ$).
- Diagonals bisect each other at right angles.
- Diagonals bisect the vertex angles.
- Opposite sides are parallel (AB || DC, AD || BC).
Step-by-Step Solution
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Consider triangle $\triangle ABC$. Since ABCD is a rhombus, sides AB and BC are equal (AB = BC). Therefore, $\triangle ABC$ is an isosceles triangle.
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In an isosceles triangle, the angles opposite the equal sides are equal. Thus, $\angle BAC = \angle BCA$. Given $\angle ACB = 40^\circ$, we have $\angle BAC = 40^\circ$.
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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 $\angle ABC$.
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In a rhombus, the diagonals bisect the angles. The diagonal BD bisects $\angle ABC$. Therefore, $\angle CBD = \frac{\angle ABC}{2} = \frac{100^\circ}{2} = 50^\circ$.
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Since opposite sides of a rhombus are parallel, AD || BC. The diagonal BD acts as a transversal line intersecting these parallel lines.
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When a transversal intersects parallel lines, alternate interior angles are equal. Therefore, $\angle ADB = \angle CBD$.
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From step 4, we know $\angle CBD = 50^\circ$. Thus, $\angle ADB = 50^\circ$.
Conclusion
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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