Let ABC be a triangle inscribed in a circle and let `l_(a)=(m_(a))/(M_(a)), l_(b)=(m_(b))/(M_(b)), l_(c )=(m_(c ))/(M_(c ))` where `m_(a), m_(b), m_(c )` are the lengths of the angle bisectors of angles A, B and C respectively , internal to the triangle and `M_(a), M_(b) and M_(c )` are the lengths of these internal angle bisectors extended until they meet the circumcircle.
Q. `l_(a)` equals :

Options

A. `(sinA)/(sin(B+(A)/(2)))`
B. `(sinBsinC)/(sin^(2)((B+C)/(2)))`
C. `(sinBsinC)/(sin^(2)(B+(A)/(2)))`
D. `(sinB+sinC)/(sin^(2)(B+(A)/(2)))`

Answer

To solve the problem, we need to find the value of \( l_a = \frac{m_a}{M_a} \), where \( m_a \) is the length of the internal angle bisector of angle \( A \) and \( M_a \) is the length of the internal angle bisector extended until it meets the circumcircle of triangle \( ABC \).

### Step-by-Step Solution:

1. **Understanding the Angle Bisector Lengths**:
The length of the internal angle bisector \( m_a \) can be calculated using the formula:
\[ m_a = \frac{2bc}{b+c} \cos\left(\frac{A}{2}\right) \]
where \( A \) is the angle at vertex \( A \), and \( b \) and \( c \) are the lengths of the sides opposite to angles \( B \) and \( C \) respectively.

2. **Finding the Extended Angle Bisector Length**:
The length of the angle bisector extended to meet the circumcircle \( M_a \) can be expressed as:
\[ M_a = \frac{2R \cdot \sin A}{\sin\left(\frac{A}{2}\right)} \]
where \( R \) is the circumradius of triangle \( ABC \).

3. **Substituting into the Ratio**:
Now we substitute \( m_a \) and \( M_a \) into the expression for \( l_a \):
\[ l_a = \frac{m_a}{M_a} = \frac{\frac{2bc}{b+c} \cos\left(\frac{A}{2}\right)}{\frac{2R \cdot \sin A}{\sin\left(\frac{A}{2}\right)}} \]

4. **Simplifying the Expression**:
We can simplify this expression:
\[ l_a = \frac{bc \cdot \cos\left(\frac{A}{2}\right) \cdot \sin\left(\frac{A}{2}\right)}{R \cdot \sin A \cdot (b+c)} \]
Using the identity \( \sin A = 2 \sin\left(\frac{A}{2}\right) \cos\left(\frac{A}{2}\right) \), we can further simplify:
\[ l_a = \frac{bc \cdot \cos\left(\frac{A}{2}\right) \cdot \sin\left(\frac{A}{2}\right)}{R \cdot 2 \sin\left(\frac{A}{2}\right) \cos\left(\frac{A}{2}\right) \cdot (b+c)} \]
Cancelling \( \cos\left(\frac{A}{2}\right) \) and \( \sin\left(\frac{A}{2}\right) \):
\[ l_a = \frac{bc}{2R(b+c)} \]

5. **Final Expression**:
Thus, the final expression for \( l_a \) is:
\[ l_a = \frac{bc}{2R(b+c)} \]