The incircle of `DeltaABC` touches side BC at D. The difference between BD and CD (R is circumradius of `DeltaABC`) is

Options

A. `|4 R sin.(A)/(2)sin.(B -C)/(2)|`
B. `|4R cos.(A)/(2) sin.(B -C)/(2)|`
C. `|b-c|`
D. `|(b-c)/(2)|`

Explanation

To solve the problem, we need to find the difference between the lengths BD and CD, where D is the point where the incircle of triangle ABC touches side BC. We will denote the lengths as follows:

- Let \( BD = x \)
- Let \( CD = y \)

The semiperimeter \( s \) of triangle ABC is given by:

\[ s = \frac{AB + AC + BC}{2} \]

From the properties of the incircle, we know:

\[ BD = s - AB \quad \text{and} \quad CD = s - AC \]

Thus, we can express the difference \( |BD - CD| \) as:

\[ |BD - CD| = |(s - AB) - (s - AC)| = |AC - AB| \]

This simplifies to:

\[ |BD - CD| = |AC - AB| \]

Next, we can relate \( |AC - AB| \) to the circumradius \( R \) of triangle ABC. We can use the sine rule and properties of triangles to express this difference in terms of \( R \).

Using the sine rule, we have:

\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]

Where \( a = BC \), \( b = AC \), and \( c = AB \). Thus, we can express \( AC \) and \( AB \) as:

\[ AC = 2R \sin B \quad \text{and} \quad AB = 2R \sin C \]

Now substituting these into our expression for \( |BD - CD| \):

\[ |BD - CD| = |2R \sin B - 2R \sin C| = 2R |\sin B - \sin C| \]

Using the sine difference identity, we can express \( |\sin B - \sin C| \):

\[ |\sin B - \sin C| = 2 \cos\left(\frac{B + C}{2}\right) \sin\left(\frac{B - C}{2}\right) \]

Substituting this back into our equation gives:

\[ |BD - CD| = 2R \cdot 2 \cos\left(\frac{B + C}{2}\right) \sin\left(\frac{B - C}{2}\right) = 4R \cos\left(\frac{B + C}{2}\right) \sin\left(\frac{B - C}{2}\right) \]

Since \( B + C = 180^\circ - A \), we have:

\[ \cos\left(\frac{B + C}{2}\right) = \cos\left(90^\circ - \frac{A}{2}\right) = \sin\left(\frac{A}{2}\right) \]

Thus, we can write:

\[ |BD - CD| = 4R \sin\left(\frac{A}{2}\right) \sin\left(\frac{B - C}{2}\right) \]

Finally, we conclude that the difference between \( BD \) and \( CD \) is:

\[ |BD - CD| = 4R \sin\left(\frac{A}{2}\right) \sin\left(\frac{B - C}{2}\right) \]