Question

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

• A. $(\sin B+\sin C):\sin A$
• B. $(\cos B+\cos C):\cos A$
• C. $\cos \left(\frac{B-C}{2}\right):\cos \left(\frac{B+C}{2}\right)$
• D. $\sin \left(\frac{B-C}{2}\right):\sin \left(\frac{B+C}{2}\right)$

Answer 1

The correct option is A $(\sin B+\sin C):\sin A$

GIVEN: A triangle $ABC$, $AI$, $BI$ & $CI$ are angle bisectors. $I$ is incentre of the triangle. $BC=a,AC=b,AB=c$

TO PROVE: $\frac{AI}{ID}=\frac{b+c}{a}=(sinB+sinC):sinA$

PROOF: By applying angle bisector theorem: which states that: Angle bisector of any angle of a triangle, bisects the opposite side in the same ratio, as that of the other $2$ sides of the triangle.

So, in triangle $ABD$,

$BI$ bisects angle $B$ ( given)

$=>\frac{c}{BD}=\frac{AI}{ID}$ ... $\left(byanglebisectortheorem\right)$ …………………..(1)

Now, in triangle $ACD$,

$CI$ bisects $\angle C$ ( given)

$=>\frac{b}{DC}=\frac{AI}{ID}$ ... $\left(bytheabovetheorem\right)$ …………………..(2)

From (1) & (2)

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

$=>\frac{DC}{BD}=\frac{b}{c}$

$=>\frac{DC+BD}{BD}=\frac{b+c}{c}$ ..$\left(byadding1toboththesidesorbylawofproportion\right)$

$=>\frac{a}{BD}=\frac{b+c}{c}$

$=>\frac{c}{BD}=\frac{b+c}{a}$

But $\frac{c}{BD}=\frac{AI}{ID}$ ...$\left(from1\right)$

Hence, $\frac{AI}{ID}=\frac{b+c}{a}=\frac{sinB+sinC}{sinA}$

$⟹AI:ID=(b+c):a=(sinB+sinC):sinA$

Hence option $A$ is the answer.