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Altitudes of ...

Question

Altitudes of a triangle are concurrent - prove by vector method.

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Solution

Let $Δ A B C$ be a triangle and let $A D, B E$ be its to altitudes intersecting at $O$ .

In order to prove that the altitudes are concurrent. it is sufficient to prove that $C O$ is perpendicular to $A B$ .

Taking $O$ as the origin, let the position vectors of $A, B, C$ be $\to a, \to b, \to c$ respectively.

Then, $\to O A = \to a; O B = \to b; O C = \to c$

Now $A D ⊥ B C$

$\Rightarrow \to O A ⊥ \to B C$

$\Rightarrow \to O A . \to O B = 0$

$\Rightarrow \to a . (\to c - \to b) = 0$

$\Rightarrow \to a . \to c - \to a . \to b = 0$ ......... (1)

$B E ⊥ C A \Rightarrow \to O B ⊥ \to C A$

$\Rightarrow \to O B . \to C A = 0 \Rightarrow \to b . (\to a - \to c) = 0$

$\Rightarrow \to b . \to a - \to b . \to c = 0$ ....... (2)

Adding (1) and (2), we get
$\to a . \to c - \to b . \to c = 0 \Rightarrow (\to a - \to b) . \to c = 0$

$\Rightarrow \to B A . \to O C = 0 \Rightarrow \to O C ⊥ \to A B$

Hence, the three altitudes are concurrent.

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