If the exterior angle of a quadrilateral formed by producing one of its sides is equal to the interior opposite angle, prove that the quadrilateral is cyclic.

Explanation

To prove that the quadrilateral ABCD is cyclic given that the exterior angle formed by extending side CD is equal to the interior opposite angle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Angles:**
Let \( \angle DCE \) be the exterior angle formed by extending side \( CD \) to point \( E \). According to the problem, we have:
\[ \angle DCE = \angle BAD \]
Let \( \angle DCE = x \). Therefore, we can write:
\[ \angle BAD = x \]

2. **Use the Linear Pair Property:**
Since \( \angle BCD \) and \( \angle DCE \) form a linear pair, we know that:
\[ \angle BCD + \angle DCE = 180^\circ \]
Substituting \( \angle DCE = x \) into the equation gives us:
\[ \angle BCD + x = 180^\circ \]
Thus, we can express \( \angle BCD \) as:
\[ \angle BCD = 180^\circ - x \]

3. **Sum of Opposite Angles:**
Now, we have two angles of the quadrilateral:
- \( \angle BAD = x \)
- \( \angle BCD = 180^\circ - x \)

To check if ABCD is cyclic, we need to prove that the sum of these two opposite angles equals \( 180^\circ \):
\[ \angle BAD + \angle BCD = x + (180^\circ - x) \]
Simplifying this gives:
\[ \angle BAD + \angle BCD = 180^\circ \]

4. **Conclusion:**
Since the sum of the opposite angles \( \angle BAD \) and \( \angle BCD \) is \( 180^\circ \), we can conclude that quadrilateral ABCD is cyclic.

### Final Statement:
Therefore, we have proved that if the exterior angle of a quadrilateral formed by producing one of its sides is equal to the interior opposite angle, then the quadrilateral is cyclic.

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