ABCD is a cyclic quadrilateral. AB and DC are the chords, when produced meet in E. Then, what kind of `Delta EBC` and `Delta EDA` are ?

Options

A. Equilateral
B. Equiangular
C. Congurent
D. None of the above

Answer

To determine the relationship between triangles \( \Delta EBC \) and \( \Delta EDA \) in the cyclic quadrilateral \( ABCD \) where \( AB \) and \( CD \) are extended to meet at point \( E \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Cyclic Quadrilaterals**:
A cyclic quadrilateral is one where all vertices lie on a single circle. A key property of cyclic quadrilaterals is that the sum of the opposite angles is \( 180^\circ \). Thus, we have:
\[ \angle BAD + \angle BCD = 180^\circ \quad \text{(Equation 1)} \]

2. **Angles on a Straight Line**:
When we extend the chords \( AB \) and \( CD \) to meet at point \( E \), the angles formed at this point must also sum to \( 180^\circ \). Therefore, we can write:
\[ \angle EAD + \angle BAD = 180^\circ \quad \text{(Equation 2)} \]

3. **Equating the Angles**:
Since both equations equal \( 180^\circ \), we can set their right-hand sides equal to each other:
\[ \angle BAD + \angle BCD = \angle EAD + \angle BAD \]
By canceling \( \angle BAD \) from both sides, we get:
\[ \angle BCD = \angle EAD \quad \text{(Equation 3)} \]

4. **Applying the Same Logic to Another Pair of Angles**:
We can apply the same reasoning to the angles \( \angle ABC \) and \( \angle ADC \):
\[ \angle ABC + \angle ADC = 180^\circ \quad \text{(Equation 4)} \]
And for the angles at point \( E \):
\[ \angle ADE + \angle ADC = 180^\circ \quad \text{(Equation 5)} \]
Again, equating the right-hand sides gives:
\[ \angle ABC + \angle ADC = \angle ADE + \angle ADC \]
Canceling \( \angle ADC \) leads to:
\[ \angle ABC = \angle ADE \quad \text{(Equation 6)} \]

5. **Conclusion About the Triangles**:
We have established that:
- \( \angle BCD = \angle EAD \)
- \( \angle ABC = \angle ADE \)

Since both pairs of angles are equal, and both triangles \( \Delta EBC \) and \( \Delta EDA \) share a common angle \( \angle E \) (the angle at point \( E \)), we can conclude that:
\[ \Delta EBC \text{ and } \Delta EDA \text{ are equiangular.} \]
Therefore, they are similar triangles.

### Final Answer:
Triangles \( \Delta EBC \) and \( \Delta EDA \) are similar triangles.