prove that some of the altitudes of a triangle is less than sum of the 3 sides of a trianglre
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given: AD BE CF are altitudes
to prove: AD+BE+CF<AC+BC+AB
proof:in triangle BCE,
angleBEC=90
Hence BC>BE(hyp is the longest side)
(1)
in triangleABD,angle ADB=90 HENCE AB>AD(hyp is the longest side)(2)
in triangleACF angle ACF=90 HENCE
AC>AF(hyp is the longest side)(3)
(1)+(2)+(3)
BC+AB+AC> BE +AD +CF
OR,
AD +BE+ CF <AB +BC +AC
HENCE PROVED!!!!
to prove: AD+BE+CF<AC+BC+AB
proof:in triangle BCE,
angleBEC=90
Hence BC>BE(hyp is the longest side)
(1)
in triangleABD,angle ADB=90 HENCE AB>AD(hyp is the longest side)(2)
in triangleACF angle ACF=90 HENCE
AC>AF(hyp is the longest side)(3)
(1)+(2)+(3)
BC+AB+AC> BE +AD +CF
OR,
AD +BE+ CF <AB +BC +AC
HENCE PROVED!!!!
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