Question

In a $△$ ABC, AD is the bisector of $\angle$ A, meeting side BC at D.

(i) If BD = 2.5 cm, AB = 5 cm and AC = 4.2 cm, find DC.

(ii) If BD = 2 cm, AB = 5 cm and DC = 3 cm, find AC.

(iii) If Ab = 3.5 cm, AC = 4.2 cm and DC = 2.8 cm, find BD.

(iv) If AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC.

(v) If AC = 4.2 cm, DC = 6 cm and BC = 10 cm, find AB.

(vi) If AB = 5.6 cm, AC = 6 cmand DC = 3 cm, find BC

(vii) If AD = 5.6 cm, BC = 6 cm and BD = 3.2 cm, find AC

(viii) If AB = 10 cm, AC = 6 cm and BC = 12 cm, find BD and DC.

Answer 1

Given in $△ABC$, $AD$ is the bisector of angle $A$

By internal angle bisector theorem, the bisector of vertical angle of a triangle divides the base in the ratio of the other two sides.

$\left(i\right)$ $\frac{AB}{AC}=\frac{BD}{DC}$

$\therefore$ $\frac{5}{4.2}=\frac{2.5}{DC}$

$\therefore$ $DC=\frac{2.5\times 4.2}{5}$

$\therefore$ $DC=2.1cm$

$\left(ii\right)$ $\frac{AB}{AC}=\frac{BD}{DC}$

$\therefore$ $\frac{5}{AC}=\frac{2}{3}$

$\therefore$ $AC=\frac{5\times 3}{2}$

$\therefore$ $AC=7.5cm$

$\left(iii\right)$ $\frac{AB}{AC}=\frac{BD}{DC}$

$\therefore$ $\frac{3.5}{4.2}=\frac{BD}{2.8}$

$\therefore$ $BD=\frac{3.5\times 2.8}{4.2}$

$\therefore$ $BD=2.33cm$

$\left(iv\right)$ $\frac{AB}{AC}=\frac{BD}{DC}$

Let $BD$ be $x$ then $DC$ becomes $6-x$

$\therefore$ $\frac{10}{14}=\frac{x}{6-x}$

$\therefore$ $\frac{5}{7}=\frac{x}{6-x}$

$\therefore$ $30-5x=7x$

$\therefore$ $12x=30$

$\therefore$ $x=2.5cm$

$\therefore$ $BD=2.5cm$ and $CD=6-2.5=3.5cm$

$\left(v\right)$ $\frac{AB}{AC}=\frac{BD}{DC}$

$\therefore$ $\frac{AB}{4.2}=\frac{10-6}{6}$

$\therefore$ $AB=\frac{4\times 4.2}{6}$

$\therefore$ $AB=2.8cm$

$\left(vi\right)$ $\frac{AB}{AC}=\frac{BD}{DC}$

$\therefore$ $\frac{5.6}{6}=\frac{BD}{3}$

$\therefore$ $BD=\frac{5.6\times 3}{6}$

$\therefore$ $BD=2.8cm$

$\Rightarrow$ $BC=BD+CD=2.8+3=5.8cm$

$\left(vii\right)$ $\frac{AB}{AC}=\frac{BD}{DC}$

$\therefore$ $\frac{5.6}{AC}=\frac{3.2}{6-3.2}$ [ AB=AD ]

$\therefore$ $AC=\frac{5.6\times 2.8}{3.2}$

$\therefore$ $AC=4.9cm$

$\left(viii\right)$ $\frac{AB}{AC}=\frac{BD}{DC}$

Let $BD$ be $x$ then $DC$ becomes $6-x$

$\therefore$ $\frac{10}{6}=\frac{x}{12-x}$

$\therefore$ $\frac{5}{3}=\frac{x}{12-x}$

$\therefore$ $60-5x=3x$

$\therefore$ $8x=60$

$\therefore$ $x=7.5cm$

$\Rightarrow$ $BD=7.5cm$ and $CD=12-7.5=4.5cm$