SOLUTION: Could you please help me solve this problem? Prove the Median Inequality: If M is the midpoint of side BC of triangle ABC, then AM is less than 1/2(AB+AC). Thank you.

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Question 448086: Could you please help me solve this problem?
Prove the Median Inequality: If M is the midpoint of side BC of triangle ABC, then AM is less than 1/2(AB+AC).
Thank you.
Found 2 solutions by richard1234, Edwin McCravy:
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
Whoops, mixed up AM with BM. It occasionally happens when I try to solve a question without pen or paper. Anyway, I have come up with a new proof, which is similar to that of the other tutor's, but uses a different method of constructing it. Note that neither proof is necessarily more "correct" than the other.

Start with the original diagram (I have used the other tutor's as a reference):

We can extend CA past A, and draw a line through B parallel to AM, and call their intersection E:


By AAA similarity, we can show that triangles EBC and AMC are similar, with a ratio of 2:1 (since BC = CM). Hence, we can establish the following ratios:

EB = 2AM
AC = AE
From the triangle inequality, EB < AB + AE. Since EB = 2AM and AC = AE, we obtain 2AM < AB + AC --> AM < (AB + AC)/2.

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!

The other tutor did not prove the theorem because
AM is not half of BC, as he stated.  Here is the correct proof:

The other tutor came back, edited it, and gave an alternate correct proof. 



Extend AM to twice its length, to D, so that AM = DM



Draw DC




       AM = DM       by construction

     ∠AMB = ∠DMC    because they are are vertical angles.

       BM = CM       because a median bisects the side it's drawn to

     ⵠABM ≅ ⵠDCM    by Side-angle-side

        AB = DC      corresponding parts of congruent triangles.

   AC + CD > AD      by triangle inequality on ⵠACD

   AC + AB > AD      substituting equals for equals.

        AD = 2AM     by construction

   AC + AB > 2AM     substituting equals for equals.

½(AC + AB) > AM      multiplying both sides by ½ .

Edwin