If AOB is a diameter of the circle and AC=BC, then `angle CAB` is equal to

Options

A. `30^(@)`
B. `60^(@)`
C. `90^(@)`
D. `45^(@)`

Explanation

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step-by-Step Solution:

1. **Identify the Given Information**:
- AOB is the diameter of the circle.
- AC = BC (meaning the lengths of chords AC and BC are equal).

2. **Draw the Circle**:
- Draw a circle with center O.
- Mark points A and B on the circumference such that AOB is the diameter.
- Mark point C on the circumference such that AC = BC.

3. **Use the Property of Isosceles Triangle**:
- Since AC = BC, triangle ACB is isosceles.
- Therefore, angles opposite to equal sides are equal. This means:
\[ \angle CAB = \angle CBA \]
- Let’s denote these angles as \( x \). Thus, we have:
\[ \angle CAB = x \quad \text{and} \quad \angle CBA = x \]

4. **Use the Property of the Diameter**:
- Since AOB is the diameter, angle ACB (the angle subtended by the diameter at the circumference) is a right angle:
\[ \angle ACB = 90^\circ \]

5. **Apply the Triangle Angle Sum Property**:
- The sum of angles in triangle ACB is equal to 180 degrees:
\[ \angle CAB + \angle CBA + \angle ACB = 180^\circ \]
- Substituting the known values:
\[ x + x + 90^\circ = 180^\circ \]
- This simplifies to:
\[ 2x + 90^\circ = 180^\circ \]

6. **Solve for x**:
- Subtract 90 degrees from both sides:
\[ 2x = 90^\circ \]
- Divide by 2:
\[ x = 45^\circ \]

7. **Conclusion**:
- Since \( x = \angle CAB \), we conclude that:
\[ \angle CAB = 45^\circ \]

### Final Answer:
\[ \angle CAB = 45^\circ \]

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