Question

In a $\Delta ABC,$ AD is the bisector of the angle A meeting BC at D. If I is the incentre of the triangle, then AI : DI is equal to -

• A. (sinB + sinC) : sinA
• B. (cosB + cosC) : cosA
• C. $cos\left(\frac{B+C}{2}\right):cos\left(\frac{B+C}{2}\right)$
• D. $cos\left(\frac{B-C}{2}\right):cos\left(\frac{B+C}{2}\right)$

Answer 1

The correct option is D $cos\left(\frac{B-C}{2}\right):cos\left(\frac{B+C}{2}\right)$

Since $AD$ is internal bisector of $\angle BAC$

$\frac{BD}{DC}=\frac{AB}{AC}=\frac{c}{b}$

$BD+DC=a$

$\Rightarrow BD=\frac{ac}{b+c}$ ; $DC=\frac{ab}{b+c}$

$BI$ is internal bisector of $\angle ABC$

$\Rightarrow \frac{AI}{ID}=\frac{AB}{BD}=\frac{c}{\frac{ac}{b+c}}=\frac{b+c}{a}$

$\Rightarrow \frac{AI}{ID}=\frac{\sin B+\sin C}{\sin A}$ ( from sine rule)

$=\frac{2\sin \frac{B+C}{2}\cos \frac{B-C}{2}}{2\sin \frac{B+C}{2}\cos \frac{B+C}{2}}=\frac{\cos \frac{B-C}{2}}{\cos \frac{B+C}{2}}$