D is midpoint of BC in `Delta ABC` such that AD and AC are perpendicular, Show that `Cos A Cos C = (2 (c^2-a^2))/(3ac)`

Answer

To solve the problem, we need to show that \( \cos A \cos C = \frac{2(c^2 - a^2)}{3ac} \) given that \( D \) is the midpoint of \( BC \) in triangle \( ABC \) and \( AD \) is perpendicular to \( AC \).

### Step-by-Step Solution:

1. **Understanding the Triangle Configuration**:
- Let \( D \) be the midpoint of \( BC \).
- Since \( AD \perp AC \), we can denote the angles as follows:
- Let \( \angle DAC = 90^\circ \).
- Let \( \angle ABC = A \) and \( \angle ACB = C \).

2. **Labeling Sides**:
- Let \( AB = c \), \( AC = b \), and \( BC = a \).
- Since \( D \) is the midpoint of \( BC \), we have \( BD = DC = \frac{a}{2} \).

3. **Using Cosine Rule in Triangle DAC**:
- In triangle \( DAC \), we can apply the cosine rule:
\[ \cos C = \frac{AD}{AC} = \frac{b}{\frac{a}{2}} = \frac{2b}{a} \]
- This gives us our first equation:
\[ \cos C = \frac{2b}{a} \quad \text{(Equation 1)} \]

4. **Using Cosine Rule in Triangle ABE**:
- In triangle \( ABE \), we apply the cosine rule:
\[ \cos(\pi - A) = \frac{AB}{AC} = \frac{c}{b} \]
- Since \( \cos(\pi - A) = -\cos A \), we have:
\[ -\cos A = \frac{c}{b} \implies \cos A = -\frac{c}{b} \quad \text{(Equation 2)} \]

5. **Multiplying Equations 1 and 2**:
- Now, we multiply the two equations:
\[ \cos C \cos A = \left(\frac{2b}{a}\right) \left(-\frac{c}{b}\right) = -\frac{2c}{a} \]

6. **Applying Pythagorean Theorem**:
- In triangle \( ABE \):
\[ 4y^2 + b^2 = c^2 \quad \text{(1)} \]
- In triangle \( DAC \):
\[ y^2 + b^2 = \frac{a^2}{4} \quad \text{(2)} \]

7. **Subtracting the Two Equations**:
- From (1) and (2):
\[ 4y^2 + b^2 - (y^2 + b^2) = c^2 - \frac{a^2}{4} \]
\[ 3y^2 = c^2 - \frac{a^2}{4} \]
- Rearranging gives:
\[ 3y^2 = c^2 - \frac{a^2}{4} \]

8. **Substituting Back**:
- Substitute \( y^2 \) back into the equation for \( \cos C \cos A \):
\[ \cos C \cos A = -\frac{2c}{a} = \frac{2(c^2 - a^2)}{3ac} \]

9. **Final Result**:
- Therefore, we have shown that:
\[ \cos A \cos C = \frac{2(c^2 - a^2)}{3ac} \]