Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
0 votes

Prove that the altitudes of a triangle are concurrent.

Can you answer this question?

1 Answer

0 votes
  • 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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App