Question

show that 3 altitude of triangle are concurrent by coordinate geometry?

Answer 1

$\huge{Hey Dude!!!}$

☆☞ Here is ur answer ☜☆

☆☞ Let ABC be any triangle.


☆☞ Let AD ⊥⊥ BC and BE ⊥⊥ to AC


☆☞ Let AD and BE intersect at O (origin say)
Join CO and extend it to meet AB at F
⇒⇒ AOD and BOE are two altitudes of the triangle.


☆☞ We have to prove that COF is the third altitude.
⇒⇒ we have to prove that CF is ⊥⊥ to AB
Let OA−→−=a→,OB−→−=b→OA→=a→,OB→=b→ and OC−→−=c→OC→=c→


☆☞ We know that AB−→−=b→−a→,BC−→−=c→−b→and AC−→−=c→−a→AB→=b→−a→,BC→=c→−b→and AC→=c→−a→


☆☞ Since AD ⊥⊥ BC and BE ⊥⊥ AC,
a→.(c→−b→)=0 and b→.(c→−a→)=0a→.(c→−b→)=0 and b→.(c→−a→)=0
⇒a→.c→=a→.b→⇒a→.c→=a→.b→..........(i) and
b→.c→=b→.a→b→.c→=b→.a→..........(ii)


☆☞ But we know that a→.b→=b→.a→a→.b→=b→.a→
⇒(i)=(ii)⇒(i)=(ii)
⇒a→.c→=b→.c→⇒a→.c→=b→.c→
⇒a→.c→−b→.c→=0⇒a→.c→−b→.c→=0
⇒(a→−b→).c→=0⇒(a→−b→).c→=0
⇒a→−b→is⊥toc→⇒a→−b→is⊥toc→
⇒AB−→−⊥ to OC−→−⇒AB→⊥ to OC→
⇒⇒ FOC is altitude of the side AB
⇒⇒ All the three altitudes meet at a common point O.

ENJOY...

GOOD BYE!!!

Answer 2

heya...

Here is your answer...

Let ∆ ABC be the given triangle.

Perpendiculars BE on AC and CF on AB intersect at O.

Construction: Join AO and produce it to meet BC at D.

Join EF.

In order to prove that the perpendicular from the vertices on opposite sides are concurrent, it is sufficient to prove that AD ⊥ BC.


BE ⊥ AC and CF ⊥ AB

⇒ ∠BEC = 90°  and ∠BFC = 90°

⇒ ∠BEC = ∠BFC = 90°

⇒ BC makes equal angles at E and F.

⇒ Points B, C, E, F are concyclic.

⇒ BCEF is a cyclic quadrilateral.

⇒ ∠ECB + ∠BFE = 180°

⇒ ∠ECB  + (∠BFC + ∠CFE)= 180°

⇒ ∠ECB + 90° + ∠CFE = 180°

⇒ ∠ACD +∠OFE = 90°  ............... (1)

In quadrilateral EOFA,

⇒ ∠OFA + OEA + = 90° + 90° = 180°

⇒ EOFA is a cyclic quadrilateral.

⇒ ∠ OFE = ∠OAE  ........... (2)  [ Angles in the same segment are equal ]

from (1) and (2)

 ∠ACD + ∠OAE = 90°

⇒ ∠ACD + ∠DAE = 90°

⇒ ∠ACD + ∠DAC = 90°

⇒ 180° – ∠ADC = 90° [ Since, in ∆ADC, ∠ACD + ∠ADC + ∠DAC = 180° ]

⇒ ∠ADC = 90°

⇒ AD ⊥ BC.

Hence altitude AD,BE and CF are concurrent.

It may help you...☺☺