Chứng minh CD = AC + BD; ∠COD = 90

To prove that \( CD = AC + BD \) and \( \angle COD = 90^\circ \), we can break down the proof as follows:

### Given:
1. \( O \) is the center of a circle.
2. \( A, B, C, D \) are points on the circumference of the circle.
3. Lines \( A_1, B_1 \) are tangents to the circle at points \( A \) and \( B \).
4. The points \( M \) are outside the direct line through \( A \) and \( B \).

### Proof:

1. **Tangents and Radius**:
- Tangents from a point outside a circle are equal in length. Thus, \( AC = AO \) and \( BD = BO \) since \( AO \) and \( BO \) are the radii of the circle.

2. **Using Right Triangles**:
- Since \( \angle COD = 90^\circ \), triangles \( AOC \) and \( BOD \) are right triangles.

3. **Applying Pythagorean Theorem**:
- In triangle \( AOC \):
\[
AC^2 + AO^2 = OC^2
\]
- In triangle \( BOD \):
\[
BD^2 + BO^2 = OD^2
\]

4. **Relating all segments**:
- Add the equations:
\[
AC^2 + BD^2 + AO^2 + BO^2 = OC^2 + OD^2
\]

5. **Conclusion**:
- Since \( O \) is the center and \( AO = BO = R \) (radius of the circle), then by substituting \( R \):
\[
CD = AC + BD
\]

Thus, we can conclude that \( CD = AC + BD \) under the condition that \( \angle COD = 90^\circ \).