Question

D, E and F are the points on sides BC, CA and AB respectively of ΔABC such that AD bisects ∠A, BE bisects ∠B and CF bisects ∠C . If AB = 5 cm, BC = 8 cm and CA = 4 cm, determine AF, CE and BD.

Answer 1

The values are $\ {AF}=\frac{5}{3} \ {cm}$ , $\ {CE}=\frac{32}{13} \ {cm}$ and $\ {BD}=\frac{40}{9} \ {cm}$

Explanation:

Given that ABC is a triangle.

Given that D,E and F are the points on sides BC, CA and AB such that AD bisects ∠A, BE bisects ∠B and CF bisects ∠C

Also given that AB = 5 cm, BC = 8 cm and CA = 4 cm

We need to determine the values of AF, CE and BD.

In ΔABC, CF bisects ∠C.

Since, the interior bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

Thus, we have,

$\frac{A F}{F B}=\frac{A C}{B C}$

Since, $\ {FB}=\ {AB}-\ {AF}$ then $\ {FB}=\ 5-\ {AF}$

Substituting the values, we get,

$\frac{A F}{5-A F}=\frac{4}{8}$

$\frac{A F}{5-A F}=\frac{1}{2}$

Cross multiplying, we get,

 $2 \mathrm{AF}=5-\mathrm{AF}$

$3 \ {AF}=5$

  $\ {AF}=\frac{5}{3} \ {cm}$

Thus, the value of AF is $\ {AF}=\frac{5}{3} \ {cm}$

Since, BE bisects ∠B , we have,

$\frac{A E}{E C}=\frac{A B}{B C}$

Since, $\ {AE}=\ {AC}-\ {CE}$ then $\ {AE}=4-\ {CE}$

Substituting the values, we have,

$\frac{4-C E}{C E}=\frac{5}{8}$

Cross multiplying, we get,

$8(4-C E)=5 C E$

$32-8 C E=5 C E$

            $32=13 \ {CE}$

          $\ {CE}=\frac{32}{13} \ {cm}$

Thus, the value of CE is $\ {CE}=\frac{32}{13} \ {cm}$

Since, AD bisects ∠A , we have,

$\frac{B D}{D C}=\frac{A D}{A C}$

Since, $\ {DC}=\ {BC}-\ {BD}=8-\ {BD}$

Substituting the values, we have,

$\frac{B D}{8-B D}=\frac{5}{4}$

Cross multiplying, we get,

$4 \ {BD}=40-5 \ {BD}$

$9 \ {BD}=40$

  $\ {BD}=\frac{40}{9} \ {cm}$

Thus, the value of BD is $\ {BD}=\frac{40}{9} \ {cm}$

Therefore, the values are $\ {AF}=\frac{5}{3} \ {cm}$ , $\ {CE}=\frac{32}{13} \ {cm}$ and $\ {BD}=\frac{40}{9} \ {cm}$

Learn more:

(1) D,E and F are the points on sites BC,CA and AB respectively of triangle ABC such that AD bisects angle A, and BE bisects angle B and CF bisect angle C . if AB = 5cm, BC = 8cm and CA = 4cm, determine AF,CE and BD.

brainly.in/question/5930678

(2) D , E and F are the points on sides BC , CA and AB respectively of △ABC . Such that AD bisects ∠A , BE bisects ∠B and CF bisects ∠C . If AB = 5cm , BC = 8cm and CA = 4cm , determine AF , CE and BD ...Draw the fig also

brainly.in/question/5173251

Answer 2

Answer is $AF = \frac {5}{3}$, $CE = \frac {13}{32}$ & $BD = \frac {40}{9}$

Step-by-step explanation:

In ABC, CF bisects $\angle C$.

Opposite Sides of the triangle gets divides in a ratio having an angle being internally divided/bisected.

Therefore,

$\frac {AF}{FB} = \frac {AC}{BC}$

⇒$\frac {AF}{5-AF} = \frac {4}{8}$ [As, FB = AB-AF= 5-AF]

⇒$\frac {AF}{5-AF} = \frac {1}{2}$

⇒ 2AF = 5-AF

⇒ 2AF +AF = 5

⇒ 3AF = 5

So, $AF = \frac {5}{3}$ cm

Again, In ABC, BE bisects $\angle B$

So, $\frac {AE}{EC} = \frac {AB}{BC}$

⇒$\frac {4-CE}{CE} = \frac {5}{8}$ [As, AE = AC-CE=4-CE]

⇒ $8(4-CE) =5 \times CE$

⇒ $32-8CE=5CE$

⇒ 32 = 13CE

⇒ $CE = \frac {13}{32}$ cm

Similarly,

⇒ $\frac {BD}{DC}=\frac {AD}{AC}$

⇒ $\frac {BD}{8-BD} = \frac {5}{4}$ [$DC=BC-BD=8-BD$]

⇒ $4BD=40-5BD$

⇒ 9BD = 40

⇒ $BD = \frac {40}{9}$ cm

Hence, $AF = \frac {5}{3}$ cm, $CE = \frac {13}{32}$ cm and $BD = \frac {40}{9}$ cm