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Prove that the perpendicular bisectors of a triangle are concurrent.

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  • To prove that the $\perp$ bisectors are concurrent, prove that the third bisector passing through the intersecting point of other two $\perp$ bisectors is also $\perp$ to its corresponding side.
  • $\overrightarrow {AB}=\overrightarrow {OB}-\overrightarrow {OA}$
  • If $\overrightarrow a.\overrightarrow b=0\:then \:\overrightarrow a\:is\:\perp\:to\:\overrightarrow b$
  • Section formula: If D is mid point of AB, then $\overrightarrow {OD}=\frac{\overrightarrow {OA}+\overrightarrow {OB}}{2}$
Let ABC be a triangle. D,E F be mid points of BC,AC and AB respectively
Let O (origin) be the point of intersection of the $\perp$ bisectors
OD and OE
Let $\overrightarrow {OA}=\overrightarrow a,\:\overrightarrow {OB}=\overrightarrow b\:and\:\overrightarrow {OC}=\overrightarrow c$
Since D is mid point of BC, we know from section formula,
$\overrightarrow {OD}=\frac{\overrightarrow {OB}+\overrightarrow {OC}}{2}=\frac{\overrightarrow b+\overrightarrow c}{2}$,
E is mid point of AC
$\Rightarrow\:\overrightarrow {OE}=\frac{\overrightarrow a+\overrightarrow c}{2}$ and
$\overrightarrow {OF}=\frac{\overrightarrow a+\overrightarrow b}{2}$
But $\overrightarrow {OD} $ is $\perp$ bisector to $\overrightarrow {BC}$
$\Rightarrow \overrightarrow {OD}.\overrightarrow {BC}=0$
$\Rightarrow\:\frac{\overrightarrow b+\overrightarrow c}{2}.(\overrightarrow c-\overrightarrow b)=0$
$\Rightarrow\:|\overrightarrow c|^2-|\overrightarrow b|^2=0$............(i)
Similarly $\overrightarrow {OE} $ is $\perp$ bisector to $\overrightarrow {AC}$
$\Rightarrow \overrightarrow {OE}.\overrightarrow {AC}=0$
$\Rightarrow\:\frac{\overrightarrow c+\overrightarrow a}{2}.(\overrightarrow c-\overrightarrow a)=0$
$\Rightarrow\:|\overrightarrow c|^2-|\overrightarrow a|^2=0$................(ii)
Subtrancting (i) - (ii) we get
$\Rightarrow\:(|\overrightarrow c|^2-|\overrightarrow b|^2)-(|\overrightarrow c|^2-|\overrightarrow a|^2)=0$
$\Rightarrow |\overrightarrow a|^2-|\overrightarrow b|^2=0$...........(iii)
To prove that the $\perp$ bisectors are concurrent, we have to prove
that the third bisector $\overrightarrow {OF}$ is $\perp$ to $\overrightarrow {AB}$
That is we have to prove that $\overrightarrow {OF}.\overrightarrow {AB}=0$
$\Rightarrow \overrightarrow {OF}.\overrightarrow {AB}=\frac{\overrightarrow a+\overrightarrow b }{2}.(\overrightarrow b-\overrightarrow a)$
$=\frac{|\overrightarrow b|^2-|\overrightarrow a|^2}{2}=0$ from (iii)
$\Rightarrow\:\overrightarrow {OF}.\overrightarrow {AB}=0$
$\Rightarrow\:\overrightarrow {OF}$ is $\perp$ to $\overrightarrow {AB}$
$\Rightarrow\:\overrightarrow {OF}$ is also a perpendicular bisector.
$\Rightarrow$ All the three $\perp$ bisectors are concurrent.
Hence proved.
answered Apr 23, 2013 by rvidyagovindarajan_1
edited Apr 24, 2013 by sreemathi.v

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