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D, E and F are the points on sides BC, CA and AB respectively of $\triangle ABC$ such that AD bisects $\angle A$, BE bisects $\angle B$ and CF bisects $\angle C$. If $AB = 5\ cm, BC = 8\ cm$ and $CA = 4\ cm$, determine AF, CE and BD.
Given:
D, E and F are the points on sides BC, CA and AB respectively of $\triangle ABC$ such that AD bisects $\angle A$, BE bisects $\angle B$ and CF bisects $\angle C$.
$AB = 5\ cm, BC = 8\ cm$ and $CA = 4\ cm$.
To do:
We have to determine AF, CE and BD.
Solution:
We know that,
The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the corresponding sides containing the angle.
Therefore,
In $\triangle ABC$, CF bisects $\angle C$.
$\frac{AF}{FB}=\frac{AC}{BC}$ $\frac{AF}{AB-AF}=\frac{4}{8}$
$\frac{AF}{5-AF}=\frac{1}{2}$ $2AF=1(5-AF)$
$2AF+AF=5$
$3AF=5$ $AF=\frac{5}{3}\ cm$
Similarly,
In $\triangle ABC$, BE bisects $\angle B$.
$\frac{AE}{EC}=\frac{AB}{BC}$ $\frac{AC-CE}{CE}=\frac{5}{8}$
$\frac{4-CE}{CE}=\frac{5}{8}$ $8(4-CE)=5(CE)$
$32-8CE=5CE$
$32=(8+5)CE$ $CE=\frac{32}{13}\ cm$
Similarly,
In $\triangle ABC$, AD bisects $\angle A$.
$\frac{BD}{DC}=\frac{AB}{AC}$ $\frac{BD}{BC-BD}=\frac{5}{4}$
$\frac{BD}{8-BD}=\frac{5}{4}$ $4BD=5(8-BD)$
$4BD=40-5BD$
$(4+5)BD=40$ $BD=\frac{40}{9}\ cm$
Therefore,
$AF=\frac{5}{3}\ cm$ , $CE=\frac{32}{13}\ cm$ and $BD=\frac{40}{9}\ cm$ .