Incentre I divides the angle bisector in a defined ratio of the length of sides of the triangle
IKAI=ab+c
ILBI=ba+c Let AN=x,BN=y,AL=p,CL=q,CK=v,BK=u.
An angle bisector of a triangle divides the opposite side into two segments that are proportional to the other two sides of the triangle.
yx=ab,x+y=c
pq=ca,q+p=b
vu=bc,u+v=a Then
x=a+bbc,y=a+bac
u=b+cac,v=b+cab
p=a+cbc,q=a+cabThe ratios of areas of triangles ANL, BKN and CLK to the area of triangle ABC are equal to 5/33 , 9/44 and 5/12 respectively.
bcpx=335
acuy=449
abvq=125 Then
a+cbc⋅a+bbc=335bc
b+cac⋅a+bac=449ac
b+cab⋅b+cab=125ab Let ab=z,ac=w. We have
zw=335(1+w)(1+z)
w=449(z+w)(1+z)
z=125(z+w)(1+w)
449(z+w)(1+z)⋅125(z+w)(1+w)=335(1+w)(1+z)
(z+w)2=916
z=34−w Substitute
w=449(34)(1+34−w)
w=117−113w
w=21
z=34−21=65
IKAI=ab+c=z+w=34
IKAI=34
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