prove that the mean of the two sides of a triangle is greater than the median through their common vertex.
Answers
How do you prove that the mean of two sides is greater than the median drawn to the common vertex?
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Terry Moore has the right idea, but let me elaborate in case it’s not clear.
You draw the triangle △ABC and let D be the midpoint of AB¯¯¯¯¯¯¯¯ . We need to prove that AC¯¯¯¯¯¯¯¯+BC¯¯¯¯¯¯¯¯2>CD¯¯¯¯¯¯¯¯ .
To “complete the parallelogram”, we basically rotate the triangle 180 degrees and add it to the bottom of the first triangle. The line CDC′ is a straight line, because the 180 degree rotation does not change the direction of the line.
How does this help? Well, by the Triangle Inequality, we have AC¯¯¯¯¯¯¯¯+AC′¯¯¯¯¯¯¯¯¯>CC′¯¯¯¯¯¯¯¯¯ (i.e. the long green line is a shorter path than the brown line and purple line combined). But CC′¯¯¯¯¯¯¯¯¯ is twice CD¯¯¯¯¯¯¯¯ , and AC′¯¯¯¯¯¯¯¯¯ is congruent to BC¯¯¯¯¯¯¯¯ . Therefore, this inequality is equivalent to AC¯¯¯¯¯¯¯¯+BC¯¯¯¯¯¯¯¯>2CD¯¯¯¯¯¯¯¯. This is equivalent to the inequality we originally wanted to prove.
Answer:
Given: ΔABC in which AD is a median.
To prove: AB + AC > 2AD.
Construction: Produce AD to E, such that AD = DE. Join EC.
Proof: In ΔADB and ΔEDC,
AD = DE (Construction)
BD = BD (D is the mid point of BC)
∠ADB = ∠EDC (Vertically opposite angles)
∴ ΔADBΔEDC (SAS congruence criterion)
⇒ AB = ED (CPCT)
In ΔAEC,
AC + ED > AE (Sum of any two sides of a triangles is greater than the third side)
∴ AC + AB > 2AD (AE = AD + DE = AD + AD = 2AD & ED = AB)
hope this will help you.
