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Prove that the altitudes of a triangle are concurrent.

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  • To prove altitudes are concurrent.draw a line through the intersection of two altitudes which is through the third vertex and prove this line is third altitude
  • Altitude is $\perp$ to the base.
  • Two vectors $\perp$ means their dot product = 0
  • $\overrightarrow {AB}=\overrightarrow {OB}-\overrightarrow {OA}$
  • $\overrightarrow a.\overrightarrow b=\overrightarrow b.\overrightarrow a$
Let ABC be any triangle.
Let AD $\perp$ BC and BE $\perp$ toAC
Let AD and BE intersect at O (origin say)
Join CO and extend it to meet AB at F
$\Rightarrow$ AOD and BOE are two altotudes of the triangle.
We have to prove that COF is the third altitude.
$\Rightarrow$ we have to prove that CF is $\perp$ to AB
Let $\overrightarrow {OA}=\overrightarrow a,\:\:\overrightarrow {OB}=\overrightarrow b$ and $\overrightarrow {OC}=\overrightarrow c$
We know that $\overrightarrow {AB}=\overrightarrow b-\overrightarrow a,\:\overrightarrow {BC}=\overrightarrow c-\overrightarrow b\:and\:\overrightarrow {AC}=\overrightarrow c-\overrightarrow a$
Since AD $\perp$ BC and BE $\perp$ AC,
$\overrightarrow a.(\overrightarrow c-\overrightarrow b)=0\:\:and\:\:\overrightarrow b.(\overrightarrow c-\overrightarrow a)=0$
$\Rightarrow\overrightarrow a.\overrightarrow c=\overrightarrow a.\overrightarrow b$..........(i) and
$\overrightarrow b.\overrightarrow c=\overrightarrow b.\overrightarrow a$..........(ii)
But we know that $ \overrightarrow a.\overrightarrow b=\overrightarrow b.\overrightarrow a$
$\Rightarrow (i)=(ii)$
$\Rightarrow \overrightarrow a.\overrightarrow c=\overrightarrow b.\overrightarrow c$
$\Rightarrow \overrightarrow a.\overrightarrow c-\overrightarrow b.\overrightarrow c=0$
$\Rightarrow (\overrightarrow a-\overrightarrow b).\overrightarrow c=0$
$\Rightarrow\:\overrightarrow a-\overrightarrow b\:is\:\perp\:to\:\overrightarrow c$
$\Rightarrow \overrightarrow {AB}\:\perp\:to\:\overrightarrow {OC}$
$\Rightarrow$ FOC is altitude of the side AB
$\Rightarrow $ All the three altitudes meet at a common point O.
answered Apr 23, 2013 by rvidyagovindarajan_1
edited Apr 23, 2013 by sreemathi.v

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