Through the centroid of an equilateral triangle, a line parallel to the base is drawn. On this line, an arbitrary point P is taken inside the triangle. Let h denote the perpendicular distance of P from the base of the triangle. Let `h_(1) and h_(2)` be the perpendicular distance of P from the other two sides of the triangle . Then :

Options

A. `h=(h_(1)+h_(2))/(2)`
B. `h=sqrt(h_(1)h_(2))`
C. `h=(2h_(1)h_(2))/(h_(1)+h_(2))`
D. `h=((h_(1)+h_(2))sqrt(3))/(4)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Setup
We have an equilateral triangle \( ABC \) with centroid \( G \). A line parallel to the base \( BC \) is drawn through \( G \), and an arbitrary point \( P \) is taken on this line inside the triangle. The distances from point \( P \) to the sides of the triangle are denoted as follows:
- \( h \): perpendicular distance from \( P \) to base \( BC \)
- \( h_1 \): perpendicular distance from \( P \) to side \( AC \)
- \( h_2 \): perpendicular distance from \( P \) to side \( AB \)

### Step 2: Area Relationships
The area of triangle \( BPC \) can be expressed in terms of \( h \):
\[ \text{Area of } \triangle BPC = \frac{1}{2} \times BC \times h \]
Since \( ABC \) is equilateral, let \( BC = a \). Therefore,
\[ \text{Area of } \triangle BPC = \frac{1}{2} \times a \times h \]

### Step 3: Area of Triangle \( ABC \)
The area of triangle \( ABC \) can be calculated as:
\[ \text{Area of } \triangle ABC = \frac{\sqrt{3}}{4} a^2 \]

### Step 4: Area of Triangle \( BGC \)
The area of triangle \( BGC \) is one-third of the area of triangle \( ABC \) because \( G \) is the centroid:
\[ \text{Area of } \triangle BGC = \frac{1}{3} \times \text{Area of } \triangle ABC = \frac{1}{3} \times \frac{\sqrt{3}}{4} a^2 \]

### Step 5: Relating Areas
Since the area of triangle \( BPC \) is equal to the area of triangle \( BGC \):
\[ \frac{1}{2} \times a \times h = \frac{1}{3} \times \frac{\sqrt{3}}{4} a^2 \]

### Step 6: Simplifying the Equation
Cancel \( a \) from both sides (assuming \( a \neq 0 \)):
\[ \frac{1}{2} h = \frac{\sqrt{3}}{12} a \]
Multiply both sides by 2:
\[ h = \frac{\sqrt{3}}{6} a \]

### Step 7: Area of Triangles \( APB \) and \( APC \)
The areas of triangles \( APB \) and \( APC \) can be expressed as:
\[ \text{Area of } \triangle APB = \frac{1}{2} \times AB \times h_1 = \frac{1}{2} a h_1 \]
\[ \text{Area of } \triangle APC = \frac{1}{2} \times AC \times h_2 = \frac{1}{2} a h_2 \]

### Step 8: Total Area of Triangle \( ABC \)
The total area of triangle \( ABC \) can also be expressed as:
\[ \text{Area of } \triangle ABC = \text{Area of } \triangle APB + \text{Area of } \triangle APC + \text{Area of } \triangle BPC \]
Substituting the areas we found:
\[ \frac{\sqrt{3}}{4} a^2 = \frac{1}{2} a h_1 + \frac{1}{2} a h_2 + \frac{1}{2} a h \]

### Step 9: Rearranging the Equation
Factoring out \( \frac{1}{2} a \):
\[ \frac{\sqrt{3}}{4} a^2 = \frac{1}{2} a (h_1 + h_2 + h) \]
Dividing both sides by \( \frac{1}{2} a \) (assuming \( a \neq 0 \)):
\[ \frac{\sqrt{3}}{2} a = h_1 + h_2 + h \]
Thus,
\[ h = \frac{\sqrt{3}}{2} a - (h_1 + h_2) \]

### Step 10: Final Relationship
From the area relationships, we can derive that:
\[ h = \frac{h_1 + h_2}{2} \]

### Conclusion
The correct relationship is:
\[ h = \frac{h_1 + h_2}{2} \]