Prove that the angle bisectors of a triangle are concurrent

Question

✓

Answer

Consider $\triangle ABC$.
Let $A D, B E$ and $C F$ be the angle bisectors of $\triangle A B C$. Let the lengths of sides $A B, B C$ and $C A$ be $z, x$ and $y$ respectively.
Let $\overline{ a }, \overline{ b }, \overline{ c }, \overline{ d }, \overline{ e }, \overline{ f }$ be the position vectors of points $A , B , C , D , E , F$ respectively.
Now, point $D$ divides $B C$ internally in the ratio $z: y$
$\therefore \overline{ d }=\frac{ yb + z \overline{ c }}{ y + z }\ldots(i)$
Point $E$ divides $C A$ internally in the ratio $x: z$
$\therefore \overline{ e }=\frac{ z \overline{ c }+x \overline{ a }}{ z +x}\ldots(ii)$
Point $F$ divides $A B$ internally in the ratio $y: x$
$\therefore \overline{ f }=\frac{x \overline{ a }+y \overline{ b }}{x+ y }\ldots(iii)$
$\therefore$ From (i), (ii), (iii), we get
$ ( y + z ) \overline{ d }= y \overline{ b }+ z \overline{ c }$
$( z +x) \overline{ e }= z \overline{ c }+x \overline{ a }$
$(x+ y ) \overline{ f }=x \overline{ a }+ y \overline{ b }$
$\therefore( y + z ) \overline{ d }+x \overline{ a }=( z +x) \overline{ e }+ y \overline{ b }=(x+ y ) \overline{ f }+ z \overline{ c }$
$=x \overline{ a }+y b + z \overline{ c } $
$\therefore \frac{(y+ z ) \overline{ d }+x \overline{ a }}{(y+ z )+x}=\frac{( z +x) \overline{ e }+y \overline{ b }}{( z +x)+y}=\frac{(x+y) \overline{ f }+ z \overline{ c }}{(x+y)+ z }=\frac{x \overline{ a }+y b + z \overline{ c }}{x+y+ z }=\overline{ p }$
This shows that the point $P$ whose position vector is $p$, lies on the three angle bisectors $A D, B E$ and $C F$ of $\triangle A B C$ dividing them in the ratios $(y+z): x,(z+x): y$ and $(x+y): z$ respectively.
Hence, the three angle bisectors are concurrent.