in triangle abc ,AD is an altitude and H is the orthocenter . prove that AH: HD= (tan B+ tan C) : Tan A

Dear Student,
Please find below the solution to the asked query:

We know thatDistance of orthocentre from vertex A is 2RcosA and from side BCis 2RcosBcosCi.e.AH=2RcosAandHD=2RcosBcosCAHHD=2RcosA2RcosBcosCAHHD=cosAcosB.cosC....iNowtanB+tanCtanA=sinBcosB+sinCcosCsinAcosA=cosA.sinB.cosC+cosB.sinCsinA.cosB.cosC=cosA.sinB+CsinA.cosB.cosC=cosA.sinπ-AsinA.cosB.cosC=cosA.sinAsinA.cosB.cosC=cosAcosB.cosCtanB+tanCtanA=cosAcosB.cosC....iiBy i and iiAHHD=tanB+tanCtanA Hence Proved

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