Find the fractional part of triangle ABC that triangle BFD occupies

Question

In the figure below, ∠ABC is bisected by BF. If AD = 3, BD = 5, and BC = 10, find the fractional part of triangle ABC that triangle BFD occupies. NOTE: No perpendicularities/altitudes are given; only the angle bisector BF and the three segment lengths.

Answer 1

here is a useful trigonometry conclusion for BF as it is drawn/described.

BF=[2*(CB*BD)(/BC+BD)]cos(B/2)

in this case BF=(20/3)cos(B/2)

know formula for BF (angle bisector line segment) because

ΔBDC=(1/2)[BD*BF+BC*BF)sin(B/2),

and

ΔBDC=(1/2)BD*BC*sin(B)=(1/2)BD*BC*2*sin(B/2)*cos(B/2)

set equal to find BF

ΔABC=(1/2)AB*BC*sin(B)=40sin(B)

ΔBDF=(1/2)BD*BF*sin(B/2)=(1/2)*5*(20/3)*cos(B/2)sin(B/2)=(1/4)*5*(20/3)sin(B)=(25/3)sin(B)

ΔBDF/ΔABC=(25/3)/40=5/24

Answer 2

FD/FC = 5/10 = 1/2 (by bisector property);

Area of ΔABC S = 1/2·8·H = 4H, where H is altitude from C of ΔABC

Area of ΔBFD s = 1/2·5·h = 2.5h, where h is altitude from F.

So, s/S = 2.5h/4H = 5h/8H. But h/H = 1/3. Because s/S = 5/8·1/3 = 5/24