Let me analyze the given question and solve it for you.

Problem:

• AD is a diameter of the circle.
• BC = CD (isosceles triangle).
• $\angle BDA = 45^\circ$.
• Find $\angle CDB$.

Solution:

Step 1: Identify the geometry rules

1. Angle in a semicircle is 90°: Since $AD$ is a diameter, $\angle BDA = 90^\circ$.
2. The triangle $BCD$ is isosceles ($BC = CD$), which means angles $\angle BCD = \angle CDB$.

Step 2: Use the given angle $\angle BDA = 45^\circ$

• The angle $\angle BDA = 45^\circ$ forms part of the right triangle $ABD$. Therefore: $\angle BDC = 90^\circ - 45^\circ = 45^\circ.$

Step 3: Divide $\angle BDC$ equally

• Since $BC = CD$, the isosceles property of triangle $BCD$ ensures that: $\angle CDB = \angle BCD = \frac{\angle BDC}{2}.$

• Substitute $\angle BDC = 45^\circ$: $\angle CDB = \frac{45^\circ}{2} = 22.5^\circ.$

Final Answer:

$\boxed{\angle CDB = 22.5^\circ.}$

Would you like additional clarification or a detailed breakdown of any part of this solution?

Related Questions:

1. What is the property of a triangle formed with the diameter of a circle as one side?
2. How does the isosceles property simplify angle calculations in triangles?
3. Why does the angle in a semicircle always measure $90^\circ$?
4. What if $\angle BDA$ was $60^\circ$—how would the answer change?
5. How does dividing an angle relate to symmetry in geometry?

Tip:

In problems involving circles and triangles, always check for symmetry (like isosceles triangles) and apply properties like "angles in a semicircle" or "equal arcs subtend equal angles."