Solve question 80 ?

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2 Answers
Oct 17, 2017

Answer is "half the". While perimeter of a triangle is greater than sum of its medians, sum of its medians is greater than half the perimeter.

Explanation:

Let us consider #DeltaABC# as shown below, where #AD# is the median which is extended to #P# so that #AD=DP# and #CP# is joined. We also have two other medians #BE# and #CF#.
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Now it is easy to prove that #DeltaABD-=DeltaPCD# as #BD=DC#, #AD=DP# and #m/_ADB=m/_CDP# and hence #AB=CP# and #AP=2AD#

Now consider #DeltaACP#. As in a triangle, sum of two sides is always greater than the third, we have #AC+CP>AP#

or #AC+AB>2AD# i.e. some of two sides is greater than twice the median to the third side.

Similarly we have #AB+BC>2BE# and #AC+BC>2CF#.

Adding the three, we get

#2(AB+BC+AC)>2(AD+BE+CF)#

or #AB+BC+AC>AD+BE+CF#

However, also consider #DeltaCFA# and observe that

#AC<AF+CF# or #AC<1/2AB+CF#

Similarly in #DeltaBFC# #BC<BF+CF# or #BC<1/2AB+CF#

adding the two we get #AC+BC<AB+2CF#

or #2CF>AC+BC-AB#

Similarly we can have #2BE>AB+BC-AC#

and #2AD>AB+AC-BC#

adding the three we get

#2(AD+BE+CF)>AB+AC+BC#

or #AD+BE+CF>1/2(AB+AC+BC)#

Hence, while perimeter of a triangle is greater than sum of its medians, sum of its medians is greater than half the perimeter.

Oct 30, 2017

jwilson.coe.uga.edu

In #DeltaABC,F,DandE# are mid points of #AB,BCandCA# respectively. So #AD,BEandCF# are three medians.

Now

  • for #DeltaABD#
    #AD+BD>AB.....[1]#

  • for #DeltaBCE#
    #BE+CE>BC.....[2]#

  • for #DeltaACF#
    #CF+AF>CA.....[3]#

Adding [1],[2],[3] we get

#(AD+BE+CF)+(BD+CE+AF)>AB+BC+CA#

#=>(AD+BE+CF)+1/2(BC+CA+AB)>AB+BC+CA#

#=>(AD+BE+CF)>1/2(AB+BC+CA)#

So

#"sum of medians">1/2"perimeter"#