Prove that, In any triangle the altitudes are concurrent.

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

answer is 1.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

In $△ A B C .$ let the altitudes $A D, B E$ meet in $O$ . Take $O$ as origin and $\bar{O A} = \bar{a}, \bar{O B} = \bar{b}, \bar{O C} = \bar{c}$

$∴ O A ⊥ B C \Rightarrow \bar{O A} \cdot \bar{B C} = 0$

$\Rightarrow \bar{O A} \cdot (\bar{O C} - \bar{O B}) = 0$

$\Rightarrow \bar{a} \cdot (\bar{c} - \bar{b}) = 0 \Rightarrow \bar{a} \cdot \bar{c} - \bar{a} \cdot \bar{b} = 0$

$\Rightarrow \bar{a} \cdot \bar{c} = \bar{a} \cdot \bar{b} . . . . . (1)$

and $O B ⊥ A C \Rightarrow \bar{O B} \cdot \bar{A C} = 0$

$\Rightarrow \bar{O B} \cdot (\bar{O C} - \bar{O A}) = 0$

$\Rightarrow \bar{b} \cdot (\bar{c} - \bar{a}) = 0 \Rightarrow \bar{b} \cdot \bar{c} - \bar{b} \cdot \bar{a} = 0$

$\Rightarrow \bar{b} \cdot \bar{c} = \bar{b} \cdot \bar{a} \Rightarrow \bar{b} \cdot \bar{c} = \bar{a} \cdot \bar{b} . . . . . (2)$

From (1) and (2) $\Rightarrow \bar{a} \cdot \bar{c} = \bar{b} \cdot \bar{c}$

$\Rightarrow (\bar{a} - \bar{b}) \cdot \bar{c} = 0 \Rightarrow (\bar{O A} - \bar{O B}) \cdot \bar{O C} = 0$

$\Rightarrow \bar{B A} \cdot \bar{O C} = 0 ∴ O C ⊥ A B$

$∴$ the line $C F$ is also an altitude

$∴$ Thus the altitudes of a $△ A B C$ are concurrent.

Watch 3-min video & get full concept clarity

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test