Answer:
m∠BCD = 80°
Step-by-step explanation:
The interior angles of a triangle sum to 180°. Therefore, in triangle ABC:
[tex]m\angle CAB + m\angle ABC + m\angle BCA = 180^{\circ} \\\\3x + 4x + 2x = 180^{\circ}[/tex]
Solve for x:
[tex]9x = 180^{\circ} \\\\\dfrac{9x}{9}=\dfrac{180^{\circ}}{9}\\\\x=20^{\circ}[/tex]
So, the measures of each angle of triangle ABC are as follows:
[tex]m\angle CAB = 3x = 3 \cdot 20^{\circ}=60^{\circ} \\\\ m\angle ABC = 4x = 4 \cdot 20^{\circ}=80^{\circ} \\\\ m\angle BCA = 2x = 2 \cdot 20^{\circ}=40^{\circ}[/tex]
Given that line segments AB and CD are parallel in the provided figure, and BC acts as a transversal intersecting these parallel lines, angles BCD and ABC are alternate interior angles.
According to the Alternate Interior Angles Theorem, when two parallel lines are intersected by a transversal, the alternate interior angles formed are congruent. Therefore, in this case, m∠BCD = m∠ABC.
Since we have already calculated that m∠ABC = 80°, then:
[tex]\LARGE\boxed{\sf m\angle BCD = 80^{\circ}}[/tex]
