MathYouLike MathYouLike
  • Toggle menubar
  • Toggle fullscreen
  • Toggle Search
  •    Sign up
  • QUESTIONS
  • TAGS
  • USERS
  • BADGES
  • UNANSWERD
  • ASK A QUESTION
  • BLOG
...

Ngô Tấn Đạt

17/09/2017 at 12:01
Answers
2
Follow

Given triangle ABC right at A ; Draw \(AH\perp BC\).Prove that : AH + BC > AB+AC

 




    List of answers
  • ...
    Kaya Renger Coordinator 17/09/2017 at 13:49

    We have :

    AH . BC = AB . AC (equals two \(S_{\Delta ABC}\))

    So 2.AH.BC = 2.AB.AC  (1)

    Apply Pythagoras theorem , we have :

    AB2 + AC2 = BC2  (2)

    But (AH + BC)2 = AH2 + BC2 + 2.AH.BC   (3)

    (AB + AC)2 = AB2 + AC2 + 2AB.AC   (4)

    From (1) ; (2) ; (3) ; (4) 

    => AH + BC > AB + AC

    Selected by MathYouLike
  • ...
    Faded 28/01/2018 at 20:43

    We have :

    AH . BC = AB . AC (equals two SΔABC

    )

    So 2.AH.BC = 2.AB.AC  (1)

    Apply Pythagoras theorem , we have :

    AB2 + AC2 = BC2  (2)

    But (AH + BC)2 = AH2 + BC2 + 2.AH.BC   (3)

    (AB + AC)2 = AB2 + AC2 + 2AB.AC   (4)

    From (1) ; (2) ; (3) ; (4) 

    => AH + BC > AB + AC


Post your answer

Please help Ngô Tấn Đạt to solve this problem!



Weekly ranking


© HCEM 10.1.29.225
Crafted with by HCEM