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 Idivides 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 ABAC and MDmeet 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 Chave 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 ABM‘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.
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