An equilateral triangle ABC is inscribed in a circle with centre O. D is a point on the minor arc BC and ∠CBD = 40°. Find the measure of ∠BCD.

The correct option is 20°.

Let us find the measure of $\angle \mathrm{BCD}$ step-by-step by utilizing the properties of equilateral triangles and cyclic quadrilaterals inscribed in a circle.

Step 1: Understand the properties of the equilateral triangle $\mathrm{ABC}$
Since triangle $\mathrm{ABC}$ is an equilateral triangle inscribed in the circle, all of its interior angles are equal to $60°$.
Therefore, we have:
$\angle \mathrm{BAC}=60°$

Step 2: Identify the cyclic quadrilateral $\mathrm{ABDC}$
The points $A$, $B$, $D$, and $C$ all lie on the circumference of the circle. Since $D$ is a point on the minor arc $\mathrm{BC}$, the order of points on the circle forms a cyclic quadrilateral, $\mathrm{ABDC}$.
A fundamental property of a cyclic quadrilateral is that the sum of its opposite angles is $180°$.
Thus, the opposite angles $\angle \mathrm{BAC}$ and $\angle \mathrm{BDC}$ must satisfy:
$\angle \mathrm{BAC}+\angle \mathrm{BDC}=180°$
Substituting $\angle \mathrm{BAC}=60°$ into the equation:
$60°+\angle \mathrm{BDC}=180°$
$\angle \mathrm{BDC}=180°-60°=120°$

Step 3: Calculate the measure of $\angle \mathrm{BCD}$ in triangle $\mathrm{BCD}$
Now consider the triangle $\mathrm{BCD}$. The sum of the interior angles in any triangle is always $180°$.
Therefore:
$\angle \mathrm{CBD}+\angle \mathrm{BDC}+\angle \mathrm{BCD}=180°$
We are given that $\angle \mathrm{CBD}=40°$ and we calculated that $\angle \mathrm{BDC}=120°$.
Substituting these values:
$40°+120°+\angle \mathrm{BCD}=180°$
$160°+\angle \mathrm{BCD}=180°$
$\angle \mathrm{BCD}=180°-160°=20°$