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