Proof: Triangle altitudes are concurrent (orthocenter) | Geometry | Khan Academy

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Showing that any triangle can be the medial triangle for some larger triangle. Using this to show that the altitudes of a triangle are concurrent (at the orthocenter).
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[@ashwinarora7561]

all the videos of khanacademy are a life saver

[@bruceedward3079]

Beautiful, simple and elegant proof, thank you

[@xedgesharpz6525]

Ok, now my question is why all the perpendicular bisectors are concurrent.....

[@yjl5373]

The meeting point is the circumcenter, and that can be proved very easily

[@4ltrz555]

@@yjl5373 he has probably found his answer by now lol

[@TechToppers]

@@4ltrz555 😂

[@muhammadjavid3681]

I would like to request the tutor to make a video on how to find and prove the orientation of vectors!

[@ch3zin]

Amazing!

[@tapaspal2968]

How is centroid of a triangle is equal to the orthocentre of medial triangle

[@Nitrodino7875]

i dont see how people dont WATCH THESE

[@akulbhendeofficial2585]

Math antics is better than this one

[@hedonism13]

The new art history videos - will they be being embedded on the KA website soon?

[@sathvikkotapati7530]

To prove the concurrency of altitudes

[@muhammadjavid3681]

If two vectors A and B are non zero vectors and A=√2B then what is the orientation of these vectors means parallel,anti parallel,coplanar and concurrent? Can anybody explain?

[@amithandge1018]

Multiplying B vector by a pure number (√2 here)will only affect the magnitude not direction so they are parallel co planar vectors

[@rajeshkumar-ec4vj]

Pls could u give the proof for concurrency of meadian s of triangle without using vectors

[@lifestyle9709]

use the ceva's theorem

[@TechToppers]

@@lifestyle9709 Or better, use midpoint theorem. It's way more elegant and pleasing.

[@SkillUpMobileGaming]

I'VE GOT A FUCKING FINAL TOMORROW...............

[@johnchristopherlicayan4065]

Hell yeah!!

[@CJhakeV]

Meh. I already know what 1+1 is..

[@sathvikkotapati7530]

But I know a simpler way

[@kevinmathewson4272]

the crime is committed at 2:35

[@dannyram1989]

Thanks for pointing that out, thought I was going crazy.

[@spiderjump]

is there another way to prove that orthocenter is concurrent?

[@TechToppers]

I know 2, vector and coordinate geometry.

[@guygremaux2512]

Illuminati