If the altitude of a triangle are in arithmetic progression, then the sides of the triangles are in

To determine the relationship between the sides of a triangle when the altitudes are in arithmetic progression (AP), we can follow these steps:

1. ### Break Down the Problem

We know that:

• Let the altitudes of the triangle be $h_1, h_2, h_3$ which are in AP.
• The sides opposite these altitudes are denoted as $a, b, c$ respectively.

2. ### Relevant Concepts

In any triangle, the area can be expressed in terms of the side and the corresponding altitude: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ From this, we can express:

1. $A = \frac{1}{2} a h_1$
2. $A = \frac{1}{2} b h_2$
3. $A = \frac{1}{2} c h_3$

3. ### Analysis and Detail

Since $h_1, h_2, h_3$ are in AP, we can write: $h_2 - h_1 = h_3 - h_2 \implies 2h_2 = h_1 + h_3 \implies h_2 = \frac{h_1 + h_3}{2}$ Using the equal areas from the definitions above, we can then set up the equations:

• From $a$ and $h_1$: $A = \frac{1}{2} a h_1$
• From $b$ and $h_2$: $A = \frac{1}{2} b \left(\frac{h_1 + h_3}{2}\right)$
• From $c$ and $h_3$: $A = \frac{1}{2} c h_3$

Equating the expressions for area gives us relationships among $a, b, c$.

4. ### Verify and Summarize

By manipulating these relationships, we would find that the sides of the triangle are proportional to the altitudes which are in an arithmetic progression, indicating that:

$\frac{a}{h_1} = \frac{b}{h_2} = \frac{c}{h_3}$

It can be concluded that if the altitudes of a triangle are in arithmetic progression, then the corresponding sides are in the same ratio.

Final Answer

If the altitudes of a triangle are in arithmetic progression, then the sides of the triangle are also in a proportional relationship, indicating that they can be considered in terms of a common ratio based on the corresponding altitudes.