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 $BD=\frac{40}{9}\ cm$ .

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Updated on: 10-Oct-2022

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