If the altitudes of a triangle are in A.P,then the sides of the triangle are in

Options

A. A.P.
B. G.P.
C. H.P.
D. none of these

Answer

To solve the problem, we need to show that if the altitudes of a triangle are in Arithmetic Progression (A.P.), then the sides of the triangle are in Harmonic Progression (H.P.).

### Step-by-Step Solution:

1. **Define the Altitudes**: Let the altitudes of the triangle from vertices A, B, and C to the opposite sides be denoted as \( X \), \( Y \), and \( Z \) respectively.

2. **Area of the Triangle**: The area \( A \) of the triangle can be expressed using the altitudes and corresponding sides:
\[ A = \frac{1}{2} \times A \times X = \frac{1}{2} \times B \times Y = \frac{1}{2} \times C \times Z \]
Here, \( A \), \( B \), and \( C \) are the lengths of the sides opposite to the vertices from which the altitudes are drawn.

3. **Equate Areas**: From the area equations, we can set:
\[ AX = BY = CZ = K \]
where \( K \) is a constant representing the area multiplied by 2.

4. **Express Altitudes in Terms of Sides**: We can express the altitudes in terms of the sides:
\[ X = \frac{K}{A}, \quad Y = \frac{K}{B}, \quad Z = \frac{K}{C} \]

5. **Use the A.P. Condition**: Since \( X, Y, Z \) are in A.P., we have:
\[ 2Y = X + Z \]
Substituting the expressions for \( X, Y, Z \):
\[ 2 \left(\frac{K}{B}\right) = \frac{K}{A} + \frac{K}{C} \]

6. **Simplify the Equation**: Dividing through by \( K \) (assuming \( K \neq 0 \)):
\[ 2 \cdot \frac{1}{B} = \frac{1}{A} + \frac{1}{C} \]

7. **Rearranging the Equation**: Rearranging gives:
\[ \frac{1}{A} + \frac{1}{C} = \frac{2}{B} \]

8. **Expressing in H.P.**: This can be rewritten as:
\[ \frac{1}{A}, \frac{1}{B}, \frac{1}{C} \text{ are in H.P.} \]
Therefore, \( A, B, C \) are in Harmonic Progression.

### Conclusion:
Thus, we conclude that if the altitudes of a triangle are in A.P., then the sides of the triangle are in H.P.

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