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