Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Use the above theorem, in the following.
If ABC is an equilateral triangle with `AD bot BC`, then prove that `AD^(2) = 3DC^(2)`.

Answer

To prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides (Pythagorean theorem), and then use this theorem to show that in an equilateral triangle \( ABC \) with \( AD \perp BC \), \( AD^2 = 3DC^2 \), we can follow these steps:

### Step 1: Prove the Pythagorean Theorem

1. **Consider a right triangle \( ABC \)** where \( \angle B \) is the right angle.
2. **Draw a perpendicular from \( B \) to \( D \)** on side \( AC \). Now we have two triangles: \( \triangle ABD \) and \( \triangle BDC \).
3. **Identify the relationships between the triangles**:
- \( \triangle ABD \sim \triangle ABC \) (by AA similarity, since they share angle \( A \) and both have a right angle).
- \( \triangle BDC \sim \triangle ABC \) (by AA similarity, since they share angle \( C \) and both have a right angle).

### Step 2: Establish Ratios from Similar Triangles

4. **From \( \triangle ABD \sim \triangle ABC \)**:
\[ \frac{AD}{AB} = \frac{AB}{AC} \]
Cross-multiplying gives:
\[ AB^2 = AD \cdot AC \quad \text{(Equation 1)} \]

5. **From \( \triangle BDC \sim \triangle ABC \)**:
\[ \frac{DC}{BC} = \frac{BC}{AC} \]
Cross-multiplying gives:
\[ BC^2 = DC \cdot AC \quad \text{(Equation 2)} \]

### Step 3: Combine the Results

6. **Add Equation 1 and Equation 2**:
\[ AB^2 + BC^2 = AD \cdot AC + DC \cdot AC \]
Factor out \( AC \):
\[ AB^2 + BC^2 = AC(AD + DC) \]

7. **Since \( AB^2 + BC^2 = AC^2 \)** (by the Pythagorean theorem):
\[ AC^2 = AC(AD + DC) \]

### Step 4: Conclude the Pythagorean Theorem

8. **This implies**:
\[ AD^2 + DC^2 = AC^2 \]
Thus, the Pythagorean theorem is proven.

### Step 5: Apply the Theorem to the Equilateral Triangle

9. **Consider equilateral triangle \( ABC \)** where \( AB = BC = AC \).
10. **Draw a perpendicular \( AD \) from vertex \( A \) to base \( BC \)**. By properties of equilateral triangles, \( D \) bisects \( BC \), so \( BD = DC \).

### Step 6: Use Pythagorean Theorem in Triangle \( ABD \)

11. **Let \( DC = x \)**, then \( BD = x \) and \( BC = 2x \).
12. **Using the Pythagorean theorem in \( \triangle ABD \)**:
\[ AD^2 + DC^2 = AC^2 \]
Since \( AC = BC = 2x \):
\[ AD^2 + x^2 = (2x)^2 \]
Simplifying gives:
\[ AD^2 + x^2 = 4x^2 \]

### Step 7: Solve for \( AD^2 \)

13. **Rearranging the equation**:
\[ AD^2 = 4x^2 - x^2 = 3x^2 \]
Therefore, we have:
\[ AD^2 = 3DC^2 \]

### Conclusion

Thus, we have proved that in an equilateral triangle \( ABC \) with \( AD \perp BC \), \( AD^2 = 3DC^2 \).