Question

In given figure, check whether AD is the bisector of $\angle$ A of $△$ ABC in each of the following:

(i) AB = 5 cm, AC = 10 cm, BD = 1.5 cm and CD = 3.5 cm

(ii) AB = 4 cm, AC = 6 cm, BD = 1.6 cm and CD = 2.4 cm

(iii) AB = 8 cm, AC = 24 cm, BD = 6 cm and BC = 24 cm

(iv) AB = 6 cm, AC = 8 cm, BD = 1.5 cm and CD = 2 cm

(v) AB = 5 cm, AC = 12 cm, BD = 2.5 cm and BC = 9 cm

Answer 1

If a line through one vertex of a triangle divides the opposite side in the ratio of the other two sides, then the line bisects the angle at the vertex.

That is if $\frac{AB}{AC}=\frac{BD}{DC}$ then $\angle BAD=\angle CAD$

1)

$AB=5cm$

$AC=10cm$

$BD=1.5cm$

$CD=3.5cm$

$\frac{AB}{AC}=\frac{5}{10}=0.5$ (1)

$\frac{BD}{CD}=\frac{1.5}{3.5}\approx 0.43$ (2)

$\left(1\right)\ne \left(2\right)$

Hence $AD$ is not the bisector of $\angle A$ of $△$ ABC

2)

$AB=4cm$

$AC=6cm$

$BD=1.6cm$

$CD=2.4cm$

$\frac{AB}{AC}=\frac{4}{6}\approx 0.67$ (1)

$\frac{BD}{CD}=\frac{1.6}{2.4}\approx 0.67$ (2)

$\left(1\right)=\left(2\right)$

Hence $AD$ is the bisector of $\angle A$ of $△$ ABC

3)

$AB=8cm$

$AC=24cm$

$BD=6cm$

$BC=24=BD+DC=6+DC$

$CD=18cm$

$\frac{AB}{AC}=\frac{8}{24}\approx 0.33$ (1)

$\frac{BD}{CD}=\frac{6}{18}\approx 0.33$ (2)

$\left(1\right)=\left(2\right)$

Hence $AD$ is the bisector of $\angle A$ of $△$ ABC

4)

$AB=6cm$

$AC=8cm$

$BD=1.5cm$

$CD=2cm$

$\frac{AB}{AC}=\frac{6}{8}=0.75$ (1)

$\frac{BD}{CD}=\frac{1.5}{2}=0.75$ (2)

$\left(1\right)=\left(2\right)$

Hence $AD$ is the bisector of $\angle A$ of $△$ ABC

5)

$AB=5cm$

$AC=12cm$

$BD=2.5cm$

$BC=9=BD+DC=2.5+DC$

$CD=6.5cm$

$\frac{AB}{AC}=\frac{5}{12}\approx 0.42$ (1)

$\frac{BD}{CD}=\frac{2.5}{6.5}\approx 0.38$ (2)

$\left(1\right)\ne \left(2\right)$

Hence $AD$ is not the bisector of $\angle A$ of $△$ ABC