# Incircle and Incenter

## Incircle and Incenter

Distance of the Incenter from the Vertex:

In triangle ABC $$\sin \frac{B}{2}=\frac{ID}{BI}=\frac{r}{BI}$$,

$$BI=\frac{r}{\sin \frac{B}{2}}$$,

$$\sin \frac{C}{2}=\frac{ID}{CI}=\frac{r}{CI}$$,

$$CI=\frac{r}{\sin \frac{C}{2}}$$,

Similarly, $$AI=\frac{r}{\sin \frac{A}{2}}$$,

Length of Tangent from Vertices to the Incircle:

In triangle $$\tan \frac{B}{2}=\frac{ID}{BD}=\frac{r}{BD}$$ ,

$$BD=\frac{r}{\tan \frac{B}{2}}$$ = s – b

BD = BF = s – b

Similarly, AF = AE = s – a

And CD = CE = s – c

Example: Let ABC be a triangle with $$\angle B={{60}^{0}}$$ . let AD be the bisector of $$\angle A$$  with D on BC. Suppose AC = 6 cm and the area of the triangle ADC is 10cm². Find the length of BD

solution:

From angle bisector theorem

$$\frac{r}{6}=\frac{p}{q}$$,

qr = 6p …(1)

Now area of triangle ABC = 10 cm²

$$=\frac{1}{2}\times DC\times AB=10$$,

$$=\frac{1}{2}\times q\times r=10$$,

$$qr=20$$ ,

From equation (1)

qr = 6p

20 = 6p

p = 20/6

=10/3