We are given an isosceles triangle $\triangle ABC$ with $AB = BC$ and angle $\angle B$ measuring 46°. The task is to find the measure of angle $\angle BCD$.

Step-by-step solution:

1. Identify the isosceles triangle: Since $AB = BC$, $\triangle ABC$ is isosceles, meaning that angles $\angle A$ and $\angle C$ are equal.

2. Sum of angles in triangle: The sum of the interior angles of any triangle is 180°. Therefore, for $\triangle ABC$, $\angle A + \angle B + \angle C = 180^\circ$ Since $\angle A = \angle C$, we can rewrite this as: $2\angle A + \angle B = 180^\circ$ Plug in $\angle B = 46^\circ$: $2\angle A + 46^\circ = 180^\circ$ Solve for $\angle A$: $2\angle A = 180^\circ - 46^\circ = 134^\circ$ $\angle A = \frac{134^\circ}{2} = 67^\circ$ So, $\angle A = \angle C = 67^\circ$.

3. Angle $\angle BCD$: Since $CD$ is a straight line, $\angle BCD$ is an exterior angle to $\triangle ABC$, and it is supplementary to $\angle C$. Therefore, $\angle BCD + \angle C = 180^\circ$ Substituting $\angle C = 67^\circ$: $\angle BCD + 67^\circ = 180^\circ$ Solving for $\angle BCD$: $\angle BCD = 180^\circ - 67^\circ = 113^\circ$

Final Answer:

The measure of $\angle BCD$ is $\boxed{113^\circ}$.

Do you want more details on any part of this explanation? Here are some related questions:

1. How do you prove that the sum of angles in a triangle is 180°?
2. Can you explain why exterior angles are supplementary to interior angles?
3. What are the properties of an isosceles triangle?
4. How do you calculate angles in other types of triangles?
5. How would the solution change if the angle $\angle B$ were different?

Tip: Always double-check whether angles form a straight line or a full circle to avoid mistakes!