Question
Prove that the angle bisectors of a triangle are concurrent.
✓
Answer
Let $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ be vertices of a triangle. Let $\mathrm{AD}, \mathrm{BE}$ and $\mathrm{CF}$ be the angle bisectors of the triangle ABD. Let $\bar{a}, \bar{b}, \bar{c}, \bar{d}, \bar{e}$ and $\bar{f}$ be the position vectors of the points $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}$ and $\mathrm{F}$ respectively. Also $\mathrm{AB}=\mathrm{zBC}=\mathrm{xAC}=\mathrm{y}$. Now, the angle bisector $\mathrm{AD}$ meets the side $\mathrm{BC}$ at the point D. Therefore, the point $\mathrm{D}$ divides the line segment $\mathrm{BC}$ internally in the ratio $\mathrm{AB}: \mathrm{AC}$, that is $z: y$. Hence, by section formula for internal division, we have $\bar{d}=\frac{z \bar{c}+y \bar{b}}{z+y}$ Similarly, we get
$\bar{e}=\frac{x \bar{a}+z \bar{c}}{x+z}$ and $\quad \bar{f}=\frac{y \bar{b}+x \bar{a}}{y+x}$
As
$
\bar{d}=\frac{z \bar{c}+y \bar{b}}{z+y}
$
$\therefore \quad(z+y) \bar{d}=z \bar{c}+y \bar{b}$
i.e. $\quad(z+y) \bar{d}+x \bar{a}=x \bar{a}+y \bar{b}+z \bar{c}$
similarly
$(x+z) \bar{e}+y \bar{b}=x \bar{a}+y \bar{b}+\mathrm{z} \bar{c}$
and $\quad(x+y) \bar{f}+z \bar{c}=x \bar{a}+y \bar{b}+\mathrm{z} \bar{c}$
$\therefore \quad \frac{(z+y) \bar{d}+x \bar{a}}{x+y+z}=\frac{(x+z) \bar{e}+y \bar{b}}{x+y+z}=\frac{(x+y) \bar{f}+z \bar{c}}{x+y+z}=\frac{x \bar{a}+y \bar{b}+z \bar{c}}{x+y+z}=\bar{h}$ (say)
Then we have
$
\bar{h}=\frac{(y+z) \bar{d}+x \bar{a}}{(y+z)+x}=\frac{(x+z) \bar{e}+y \bar{b}}{(x+z)+y}=\frac{(x+y) \bar{f}+z \bar{c}}{(x+y)+z}
$
That is point $\mathrm{H}(\bar{h})$ divides $\mathrm{AD}$ in the ratio $(y+z): x, \mathrm{BE}$ in the ratio $(x+z): y$ and $\mathrm{CF}$ in the ratio $(x+y): z$.
This shows that the point $\mathrm{H}$ is the point of concurrence of the angle bisectors $\mathrm{AD}, \mathrm{BE}$ and $\mathrm{CF}$ of the triangle $\mathrm{ABC}$, Thus, the angle bisectors of a triangle are concurrent and $\mathrm{H}$ is called incentre of the triangle $\mathrm{ABC}$.
$\bar{e}=\frac{x \bar{a}+z \bar{c}}{x+z}$ and $\quad \bar{f}=\frac{y \bar{b}+x \bar{a}}{y+x}$
As
$
\bar{d}=\frac{z \bar{c}+y \bar{b}}{z+y}
$
$\therefore \quad(z+y) \bar{d}=z \bar{c}+y \bar{b}$
i.e. $\quad(z+y) \bar{d}+x \bar{a}=x \bar{a}+y \bar{b}+z \bar{c}$
similarly
$(x+z) \bar{e}+y \bar{b}=x \bar{a}+y \bar{b}+\mathrm{z} \bar{c}$
and $\quad(x+y) \bar{f}+z \bar{c}=x \bar{a}+y \bar{b}+\mathrm{z} \bar{c}$
$\therefore \quad \frac{(z+y) \bar{d}+x \bar{a}}{x+y+z}=\frac{(x+z) \bar{e}+y \bar{b}}{x+y+z}=\frac{(x+y) \bar{f}+z \bar{c}}{x+y+z}=\frac{x \bar{a}+y \bar{b}+z \bar{c}}{x+y+z}=\bar{h}$ (say)
Then we have
$
\bar{h}=\frac{(y+z) \bar{d}+x \bar{a}}{(y+z)+x}=\frac{(x+z) \bar{e}+y \bar{b}}{(x+z)+y}=\frac{(x+y) \bar{f}+z \bar{c}}{(x+y)+z}
$
That is point $\mathrm{H}(\bar{h})$ divides $\mathrm{AD}$ in the ratio $(y+z): x, \mathrm{BE}$ in the ratio $(x+z): y$ and $\mathrm{CF}$ in the ratio $(x+y): z$.
This shows that the point $\mathrm{H}$ is the point of concurrence of the angle bisectors $\mathrm{AD}, \mathrm{BE}$ and $\mathrm{CF}$ of the triangle $\mathrm{ABC}$, Thus, the angle bisectors of a triangle are concurrent and $\mathrm{H}$ is called incentre of the triangle $\mathrm{ABC}$.
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