Question

If a,b and c are the sides of triangle ABC and I is the incentre of the circle inscribed in the triangle ABC and AI cuts BC at P, then AI=?
(A)[(b+c)/(a+b+c)]×AP
(B)[(c+a)/(a+b+c)]×AP
(C)[(a+b)/(a+b+c)]×AP
(D)[a/(a+b+c)]×AP
​​​​​​

Answer 1

Incenter of a triangle is the meeting point of angle bisectors of triangle.
So AI, BI & CI are the angle bisectors of triangle ABC.

Considering the triangle ABP, BI is the bisector of angle B.
So by property of angle bisector of a triangle [It divides the base in the same ratio as that of the two sides containing it],
PI/AI = BP/c;

Similarly from the other triangle AIC, PI/AI = PC/b

==> PI/AI = BP/c = PC/b

==> PI/AI = (BP + PC)/(c + b) = BC/(b + c) = a/(b + c)

Adding one on both sides,
AP/AI = (a + b + c)/(b + c)

Hence, AI/AP = (b + c)/(a + b + c)
hence AI= ((b+c)/(a+b+c))AP