In isosceles triangle ABC, sides AB and AC are equal. If point D lies in base BC and point E lies on BC produced (BC being produced through vertex C), prove that :
(i) `AC gt AD`
(ii) `AE gt AC`
(iii) `AE gt AD`

Answer

To prove the statements regarding the isosceles triangle ABC, we will follow a systematic approach using geometric properties and the Pythagorean theorem.

### Step-by-Step Solution

1. **Draw the Triangle and Points**:
- Draw triangle ABC where AB = AC (isosceles triangle).
- Let D be a point on the base BC and E be a point on the extension of BC beyond C.

2. **Identify the Perpendicular**:
- Draw a perpendicular line from point A to line BC, meeting at point F. This will help in applying the Pythagorean theorem.

3. **Establish Equal Segments**:
- Since triangle ABC is isosceles, the perpendicular from A to BC bisects BC at point F. Therefore, we have BF = FC.

4. **Apply Pythagorean Theorem in Triangle AFB**:
- In triangle AFB, by the Pythagorean theorem:
\[ AB^2 = AF^2 + BF^2 \tag{1} \]

5. **Apply Pythagorean Theorem in Triangle AFD**:
- In triangle AFD, we have:
\[ AD^2 = AF^2 + DF^2 \tag{2} \]

6. **Apply Pythagorean Theorem in Triangle ACF**:
- In triangle ACF, we have:
\[ AC^2 = AF^2 + CF^2 \tag{3} \]
- Since BF = FC, we can substitute CF with BF in equation (3):
\[ AC^2 = AF^2 + BF^2 \]

7. **Compare AC and AD**:
- From equations (1) and (2), since BF > DF (D is between B and C), we can conclude:
\[ AC^2 > AD^2 \implies AC > AD \tag{4} \]

8. **Apply Pythagorean Theorem in Triangle AFE**:
- In triangle AFE, we have:
\[ AE^2 = AF^2 + EF^2 \tag{5} \]
- Since E lies on the extension of BC, EF > BF.

9. **Compare AE and AC**:
- From equations (3) and (5), since EF > BF:
\[ AE^2 > AC^2 \implies AE > AC \tag{6} \]

10. **Compare AE and AD**:
- Since EF > DF (E is beyond D on line BC):
\[ AE^2 > AD^2 \implies AE > AD \tag{7} \]

### Conclusion
From the above steps, we have proven:
1. \( AC > AD \)
2. \( AE > AC \)
3. \( AE > AD \)