Question

AM is a median of $\Delta ABC$ . Prove that $AB+BC+CA > 2AM$

Answer 1

Hint: Start by drawing a neat diagram. Then apply the theorem that the sum of two sides of a triangle is always greater than the third side in $\Delta AMB$ and $\Delta AMC$ separately, such that AM is the third side in both the cases. Add the two inequalities to get the answer.

Complete step by step solution:
To start with the solution, let us first draw the diagram.

Let us start the solution to the above question by drawing some results related to $\Delta ABM$ . We know that the sum of two sides of a triangle is always greater than the third side, so we can say that:
$AB+BM > AM........(i)$
Similarly, if we take up $\Delta ACM$ and use the theorem that the sum of two sides of a triangle is always greater than the third side, we get
$AC+CM > AM........(ii)$
Now we will add inequality (i) and inequality (ii). On doing so, we get
$AC+CM+AB+BM > AM+AM$
$\Rightarrow AC+CM+AB+BM > 2AM$
Now if we look in the figure, we will find that BM+CM=BC. If we use this in our inequality, we get
$AB+BC+CA > 2AM$
Hence, we can say that we have proved the inequality asked in the question.

Note: The key to such questions is the diagram. If you get the diagram correct, then there is a high possibility that you would reach an answer provided you have the basic knowledge of regularly used theorems, as we saw in the above question. Also, the general mistake that a student makes is getting confused between the medians and the angle bisectors leading to errors, so be careful while using the properties of medians and angle bisectors. However, the above theorem is not specific to medians and can be proved for any line joining A and a point M on the line segment BC.