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sreesuja
Average Member

India
13 Posts

Posted - 11/17/2013 :  23:46:14  Show Profile  Reply with Quote
PLEASE HELP ME OUT

SUM OF THE ALTITUDES OF TRIANGLE IS LESS THAN HE THE SUM OF THE SIDES OF THE TRIANGLE.
(AE+CF+BD < AB+BC+CD)


PROOF
I HAVE TRIED WITH
SUM OF THE TWO SIDES IS GREATER THAN THE THIRD SIDE
ABC AS SIDES OF THE TRIANGLE AND AE. BD AND CF AS THE ALTITUDE.
I GOT

AE+EC>AC : BD+DC>BC : CF+AF>AC
AE+BE>AB : BD+AD>AB : CF+FB>BC

ADDING ALL I GOT 2AE+2BD+2CF>AB+BC+CA
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the_hill1962
Advanced Member

USA
1468 Posts

Posted - 11/18/2013 :  18:31:21  Show Profile  Reply with Quote
Your proof is interesting. It shows that AE+BD+CF > 0.5(AB+BC+AC).
However, as you find, this doesn't prove AE+CF+BD < AB+BC+CD.
Here is a hint as to prove what you are asking:
See that the triangles formed are right triangles and use the Pythagorean theorem to show that a leg (which are the altitudes) is always less than the hypotenuse.
Once you show that, you can say that the sum of all three of those "legs" is less than the sum of all three of the "hypotenuses".
If you need further help, please let us know.
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