Question

The perimeter of $△ABC$ is 49 cm. The angle bisector of $\angle A$ meets BC at M. BM = 6 cm, MC = 8 cm. Find AC.

Answer 1

Given perimeter = 49 cm

BC = BM + MC = 6 + 8 = 14 cm
AB + AC = 49 - 14 = 35 cm

Since AM is the angle bisector of $\angle A$,

$\frac{AB}{AC}=\frac{BM}{MC}$

$\Rightarrow \frac{8}{6}=\frac{3}{4}$ (1 mark)

Adding 1 on both sides, we have

$\frac{AB}{AC}+1=\frac{3}{4}+1$

$\Rightarrow \frac{AB+AC}{AC}=\frac{7}{4}$

$\Rightarrow \frac{35}{AC}=\frac{7}{4}$

$\Rightarrow AC=\frac{35\times 4}{7}$

= 20 cm (1 mark)

AB = 35 - AC

= 35 - 20

= 15 cm (1 mark)

AC = $\frac{4}{3}$$\times$ AB

= $\frac{4}{3}$$\times$ 15 cm

= 20 cm (1 mark)