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## Friday, 8 August 2014

### CHAPTER 13- Worked Out Examples –3

 Example: 10
 Show by vector methods that the angular bisectors of a triangle are concurrent and find the position of the point of concurrency in terms of the position vectors of the vertices.
 Solution: 10
Let the vertices of the triangle by $A\left( {\vec a} \right),\,\,B\,(\vec b)$ and $C(\vec c)$. We use the geometrical fact that an angle bisector divides the opposite side in the ratio of the sides containing the angle.
We thus have,
 $\dfrac{{BD}}{{DC}} = \dfrac{c}{b}$
Thus, $D$ is given by (the internal division formula):
 $D \equiv \dfrac{{z\vec c + y\vec b}}{{z + y}}$
In $\Delta ABD$, since $BI$ is the angle bisector, we have
 $\dfrac{{DI}}{{IA}} = \dfrac{{BD}}{{BA}} = \dfrac{{\dfrac{z}{{z + y}}x}}{z} = \dfrac{x}{{z + y}}$
Thus, we now have the position vectors of $A$ and $D$ we know what ratio $I$divides $AD$ in. $I$ can now be easily determined using the internal division formula:
 $I \equiv \dfrac{{x\vec a + \left( {x + y} \right)\left( {\dfrac{{z\vec c + y\vec b}}{{z + y}}} \right)}}{{x + z + y}}$ $= \dfrac{{x\vec a + y\vec b + z\vec c}}{{x + y + z}}$ $\ldots(1)$
The symmetrical nature of this expression proves that the bisector of $C$ will also pass through $I$. The angle bisectors will therefore be concurrent at $I$, called the incentre. The position vector of the incentre is given by $(1).$
 Example: 11
 In a parallelogram $ABCD$, let $M$ be the mid-point of $AB$. $AC$ and $MD$meet in $E$. Prove that both $AC$ and $MD$ are trisected at $E$.
 Solution: 11
There’s no loss of generality in assuming $A$ to be the origin $(\vec 0)$. Let $B$ and $C$have the position vectors $\vec b$ and $\vec c$. The position vector of $D$ is given by $\overrightarrow {AD}$where
 $\overrightarrow {AD} = \overrightarrow {AC} + \overrightarrow {CD}$ $= \vec c - \overrightarrow {AB}$ $= \overrightarrow c - \overrightarrow b$
$D$ is therefore the point $\overrightarrow c - \overrightarrow b$.
Since $M$ is the mid-point of $AB$$M$‘s position vector is $\dfrac{{\vec b}}{2}$. Let us first find the position vector of a point $E$‘ which lies on $AC$ and trisects it. We’ll then show that same point lies on $DM$ and trisects on $DM$ and trisects it too, proving the stated assertion.
 $E' \equiv \dfrac{{1 \times \vec c + 2 \times \vec 0}}{3} = \dfrac{{\vec c}}{3}$
A point $E$” which trisects $MD$ is
 $E' \equiv \dfrac{{2 \times \dfrac{{\vec b}}{2} + 1 \times \left( {\vec c - \vec b} \right)}}{3} = \dfrac{{\vec c}}{3}$
Since $E$‘ and $E$” are the same-point, say $E$, we see that $AC$ and $MD$ are trisected at $E$.