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

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  • Опубликовано: 14 янв 2025

Комментарии • 29

  • @ashwinarora7561
    @ashwinarora7561 12 лет назад +10

    all the videos of khanacademy are a life saver

  • @xedgesharpz6525
    @xedgesharpz6525 11 лет назад +12

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

    • @yjl5373
      @yjl5373 6 лет назад +2

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

    • @4ltrz555
      @4ltrz555 4 года назад +7

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

    • @TechToppers
      @TechToppers 3 года назад

      @@4ltrz555 😂

    • @sourthakjackmcober29
      @sourthakjackmcober29 3 месяца назад

      One of the regretful comments to have. But it is what many would wonder. Nothing wrong

  • @bruceedward3079
    @bruceedward3079 7 лет назад +5

    Beautiful, simple and elegant proof, thank you

  • @muhammadjavid3681
    @muhammadjavid3681 6 лет назад +1

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

  • @ch3zin
    @ch3zin 13 лет назад +3

    Amazing!

  • @hedonism13
    @hedonism13 13 лет назад

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

  • @tapaspal2968
    @tapaspal2968 5 лет назад +1

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

    • @Z-eng0
      @Z-eng0 Месяц назад

      Nah, it's that the circumcentre of a triangle is the same as the orthocentre of the medial triangle.
      Which is rather curious in my opinion, and I was rather shocked by that, I never knew how I might go about showing that a triangle's altitudes are concurrent, considering that altitudes felt like those vaguely defined/drawn lines that are only parallel to the opposite side or its extension, felt too vague so didn't expect the proof to be this straightforward

  • @Nitrodino7875
    @Nitrodino7875 10 лет назад +1

    i dont see how people dont WATCH THESE

  • @rajeshkumar-ec4vj
    @rajeshkumar-ec4vj 7 лет назад

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

    • @lifestyle9709
      @lifestyle9709 4 года назад +1

      use the ceva's theorem

    • @TechToppers
      @TechToppers 3 года назад

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

  • @sathvikkotapati7530
    @sathvikkotapati7530 4 года назад

    To prove the concurrency of altitudes

  • @muhammadjavid3681
    @muhammadjavid3681 6 лет назад

    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
      @amithandge1018 Год назад

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

  • @akulbhendeofficial2585
    @akulbhendeofficial2585 3 месяца назад +1

    Math antics is better than this one

  • @SkillUpMobileGaming
    @SkillUpMobileGaming 8 лет назад +2

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

  • @kevinmathewson4272
    @kevinmathewson4272 7 лет назад

    the crime is committed at 2:35

    • @dannyram1989
      @dannyram1989 5 лет назад

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

  • @CJhakeV
    @CJhakeV 13 лет назад +1

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

  • @sathvikkotapati7530
    @sathvikkotapati7530 4 года назад

    But I know a simpler way

  • @guygremaux2512
    @guygremaux2512 10 лет назад

    Illuminati

  • @spiderjump
    @spiderjump 8 лет назад

    is there another way to prove that orthocenter is concurrent?

    • @TechToppers
      @TechToppers 3 года назад

      I know 2, vector and coordinate geometry.