Question

prove that three altitudes of triangle passes through same point

Answer 1

They are going to be concurrent becauz of any triangle i can make it the medical triangle of a larger one and then its altitudes will be the perpendicular bisector for larger triangle and we already know tha perpendicular bisectors of any triangle are concurrent. They do intersect at exactly same point..

Answer 2

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HERE IS YOUR ANSWER##

Let ABC be any triangle.

Let AD ⊥ BC and BE ⊥ toAC

Let AD and BE intersect at O (origin say)

Join CO and extend it to meet AB at F

⇒ AOD and BOE are two altotudes 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 and →OC=→c

We know that →AB=→b−→a,→BC=→c−→band→AC=→c−→a

Since AD ⊥ BC and BE ⊥ AC,

→a.(→c−→b)=0and→b.(→c−→a)=0

⇒→a.→c=→a.→b..........(i) and

→b.→c=→b.→a..........(ii)

But we know that →a.→b=→b.→a

⇒(i)=(ii)

⇒→a.→c=→b.→c

⇒→a.→c−→b.→c=0

⇒(→a−→b).→c=0

⇒→a−→bis⊥to→c

⇒→AB⊥to→OC

⇒ FOC is altitude of the side AB

⇒ All the three altitudes meet at a common point O.


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