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H is the orthocentre of triangle ABC. Prove that
BC/AH+CA/BH+AB/CH≥3√3
plz upload actual question, because this question needs modification
considering the question to be find the value of the given expression,
BC/AH+CA/BH+AB/CH
here BC/AH=(AD/AH)(DC/AD)(BC/DC)
=(sinB)(sinA/cosA)(1/sinB)
=tanA
same way CA/BH=tanB
AB/CH=tanC
or, tanA+tanB+tanC=tanAtanBtanC
[where 0=tan(A+B+C)=(tanA+tanB+tanC-tanAtanBtanC)/(1-tanAtanB-tanAtanC-tanBtanC)]
=(BC/AH)(CA/BH)(AB/CH)
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