Let's solve the problem step by step.

Given:

• $\overline{BD}$ is a line intersecting the diameter $\overline{AC}$ of the circle at point $X$.
• $\angle BCA = 26^\circ$
• $\angle CAD = 47^\circ$

We are asked to find the angle $\angle BXD$.

Approach:

1. Use the property of angles in a circle: The angle subtended by the diameter on the circumference is always $90^\circ$. Therefore, $\angle BAC = 90^\circ$.

2. Apply the sum of angles in triangle $ABC$: In triangle $ABC$, $\angle BAC + \angle BCA + \angle ABC = 180^\circ$ Substitute the known angles: $90^\circ + 26^\circ + \angle ABC = 180^\circ$ So, $\angle ABC = 180^\circ - 116^\circ = 64^\circ$

3. Use the given angle $\angle CAD$: We know $\angle CAD = 47^\circ$. We can use this along with $\angle ABC$ to find $\angle BXD$.

4. Calculate $\angle BXD$: Since $\overline{BD}$ is a chord intersecting the diameter at $X$, the angles subtended by $\overline{BD}$ on the opposite sides of $\overline{AC}$ are equal. Hence, $\angle BXD = \angle ABC + \angle CAD$.

   Thus, $\angle BXD = 64^\circ + 47^\circ = 111^\circ$

So, the angle $\angle BXD = 111^\circ$.

Would you like to go over the steps in more detail or have any questions?

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Related Questions:

1. What is the significance of the angle subtended by the diameter on the circumference?
2. How would the solution change if $\overline{BD}$ did not intersect at the diameter?
3. How do you find angles in cyclic quadrilaterals?
4. What are the properties of angles formed by intersecting chords in a circle?
5. Can the angle $\angle BXD$ be more than $180^\circ$? Why or why not?
6. How can you determine if $\overline{BD}$ is a tangent to the circle?
7. What other methods could be used to solve for $\angle BXD$?
8. What if $\angle BCA$ were not given? How would you approach the problem?

Tip:

When working with angles in circles, always consider properties like the angle subtended by the diameter, angles in the same segment, and cyclic quadrilaterals to simplify your calculations.