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H is the orthocentre of triangle ABC and RR is the circumradius. Prove that, 2R<AH+BH+CH≤3R
AH=2RcosA, BH=2RcosB, CH=2RcosC
(AH+BH+CH)=2R(cosA+cosB+cosC)
=2R(1+r/R)
=2(R+r)
<=2R+R since r<=R/2 for any triangle
=3R
2R<=(AH+BH+CH)<=3R
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