Answer to Question #146223 in Geometry for ankit

Question

Let AK, BL, CN be angle bisectors of triangle ABC, and they intersect at point I. It is known that the ratios of areas of triangles BKN and CLK to the area of triangle ABC are equal to \frac{1}{10} and\frac{16}{25} respectively, and ratio IK:AI is equal to \frac{4}{5}. Find the ratio of the area of triangle ANL to the area of triangle ABC.

Answer 1

Let`s start with recognising given valuables:

a+b+c=180⁰

This means that all angles in the inside of triangle equals to 180 degrees.

S_BNK=S_ABC/10,

S_KLC=(S_ABC*16)/25.

Let`s make some substitutions:

S_ANO=5n, S_NOK=4n, S_ALO=5m, S_LOK=4m, S_ABC=S

This means that the angle bisector divides the triangle in the same ratio as it divides itself.

Then:

S_ANO+S_NOK+ S_ALO+S_LOK=S_ABC-S_BNK-S_KLC.

9n+ 9m=S-S/10-16*S/25 ==> 9*(m+n)=(13*S)/(50*9),

S_ANL=5*m+5*n=5*(m+n)= (5*13*S)/(50*9),

S_ANL=13*S_ABC/90

S_ANL/S_ABC=13/90 .

Answer: 13/90 .