Let AD, BE, CF be the lengths of internal bisectors of angles A, B, C respectively of triangle ABC. Then the harmonic mean of `AD"sec"(A)/(2),BE"sec"(B)/(2),CF"sec"(C )/(2)` is equal to :

Options

A. Harmonic mean of sides of `DeltaABC`
B. Geometric mean of sides of `DeltaABC`
C. Arithmetic mean of sides of `DeltaABC`
D. Sum of reciprocals of the sides of `DeltaABC`

Answer

To solve the problem, we need to find the harmonic mean of the expressions \(AD \sec \left(\frac{A}{2}\right)\), \(BE \sec \left(\frac{B}{2}\right)\), and \(CF \sec \left(\frac{C}{2}\right)\), where \(AD\), \(BE\), and \(CF\) are the lengths of the internal bisectors of angles \(A\), \(B\), and \(C\) respectively in triangle \(ABC\).

### Step-by-Step Solution:

1. **Find the Length of the Internal Bisector \(AD\)**:
The length of the internal bisector \(AD\) can be calculated using the formula:
\[ AD = \frac{2bc}{b+c} \cos\left(\frac{A}{2}\right) \]
Thus, we have:
\[ AD \sec\left(\frac{A}{2}\right) = \frac{2bc}{b+c} \]

2. **Find the Length of the Internal Bisector \(BE\)**:
Similarly, for the internal bisector \(BE\):
\[ BE = \frac{2ac}{a+c} \cos\left(\frac{B}{2}\right) \]
Therefore:
\[ BE \sec\left(\frac{B}{2}\right) = \frac{2ac}{a+c} \]

3. **Find the Length of the Internal Bisector \(CF\)**:
For the internal bisector \(CF\):
\[ CF = \frac{2ab}{a+b} \cos\left(\frac{C}{2}\right) \]
Hence:
\[ CF \sec\left(\frac{C}{2}\right) = \frac{2ab}{a+b} \]

4. **Calculate the Harmonic Mean**:
The harmonic mean (HM) of three quantities \(x_1\), \(x_2\), and \(x_3\) is given by:
\[ HM = \frac{3}{\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3}} \]
Applying this to our expressions:
\[ HM = \frac{3}{\frac{b+c}{2bc} + \frac{a+c}{2ac} + \frac{a+b}{2ab}} \]

5. **Simplify the Expression**:
Combine the fractions in the denominator:
\[ HM = \frac{3}{\frac{(b+c)a + (a+c)b + (a+b)c}{2abc}} \]
This simplifies to:
\[ HM = \frac{3 \cdot 2abc}{(b+c)a + (a+c)b + (a+b)c} \]

6. **Recognize the Denominator**:
The denominator can be rewritten as:
\[ (b+c)a + (a+c)b + (a+b)c = ab + ac + ba + bc + ca + cb = 2(ab + ac + bc) \]
Therefore, we have:
\[ HM = \frac{3 \cdot 2abc}{2(ab + ac + bc)} = \frac{3abc}{ab + ac + bc} \]

### Final Answer:
The harmonic mean of \(AD \sec\left(\frac{A}{2}\right)\), \(BE \sec\left(\frac{B}{2}\right)\), and \(CF \sec\left(\frac{C}{2}\right)\) is:
\[ HM = \frac{3abc}{ab + ac + bc} \]