In a `DeltaABC`, with usual notations, if `b gt c` then distance between foot of median and foot of altitude both drawn from vertex A on BC is :

Options

A. `(a^(2)-b^(2))/(2c)`
B. `(b^(2)-c^(2))/(2a)`
C. `(b^(2)+c^(2)-a^(2))/(2a)`
D. `(b^(2)+c^(2)-a^(2))/(2c)`

Answer

To find the distance between the foot of the median and the foot of the altitude drawn from vertex A on side BC in triangle ABC, where \( b > c \), we can follow these steps:

### Step-by-Step Solution:

1. **Draw Triangle ABC**: Start by sketching triangle ABC with vertices A, B, and C. Label the sides opposite to these vertices as \( a \) (opposite A), \( b \) (opposite B), and \( c \) (opposite C).

2. **Identify the Foot of the Median and the Foot of the Altitude**:
- Let D be the foot of the median from A to BC. The median divides side BC into two equal segments, so \( BD = DC \).
- Let E be the foot of the altitude from A to BC. This means AE is perpendicular to BC.

3. **Express Lengths**:
- Since D is the midpoint of BC, we can denote \( BD = DC = \frac{a}{2} \).
- The total length of BC is \( a \) (i.e., \( BC = a \)).

4. **Use the Cosine Rule to Find EC**:
- The length \( EC \) can be expressed using the cosine of angle C. From triangle AEC, we can use the cosine rule:
\[ EC = b \cos C \]
- The cosine of angle C can be calculated using the formula:
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]
- Thus, substituting this into our expression for EC gives:
\[ EC = b \cdot \frac{a^2 + b^2 - c^2}{2ab} = \frac{a^2 + b^2 - c^2}{2a} \]

5. **Find DE**:
- The distance DE between the foot of the median (D) and the foot of the altitude (E) can be calculated as:
\[ DE = BD - EC \]
- We know \( BD = \frac{a}{2} \) and substituting \( EC \) gives:
\[ DE = \frac{a}{2} - \frac{a^2 + b^2 - c^2}{2a} \]

6. **Simplify DE**:
- To simplify, we can combine the fractions:
\[ DE = \frac{a}{2} - \frac{a^2 + b^2 - c^2}{2a} = \frac{a^2 - (a^2 + b^2 - c^2)}{2a} \]
- This simplifies to:
\[ DE = \frac{c^2 - b^2}{2a} \]

### Final Result:
Thus, the distance between the foot of the median and the foot of the altitude from vertex A on BC is:
\[ DE = \frac{c^2 - b^2}{2a} \]