Question

In \[\Delta ABC\], if AD, BE and CF are the three medians, then which of the following is always the correct conclusion?
A) \[AB + BC + AC < AD + BE + CF\]
B) \[AB + BC + AC > AD + BE + CF\]
C) \[AB + BC + AC < \dfrac{2}{3}\left( {AD + BE + CF} \right)\]
D) \[AB + BC + AC = 2\left( {AD + BE + CF} \right)\]

Answer 1

Hint:
First, we will use the sum of two sides is greater than two times the median on the third side for the sides AB and AC and AD is the median on the third side, sides AB and BC and BE is the median on the third side and sides AC and BC and CF is the median on the third side. Then we will add the obtained equations to find the required inequality.

Complete step by step solution:
We are given that \[\Delta ABC\] is a triangle and AD, BE and CF are the three medians.

We know that the sum of two sides is greater than two times the median on the third side.
Since we have the sides AB and AC and AD is the median on the third side, we get
\[ \Rightarrow AB + AC > 2AC{\text{ ......eq.(1)}}\]
We know the sides AB and BC and BE is the median on the third side, we get
\[ \Rightarrow AB + BC > 2BE{\text{ ......eq.(2)}}\]
We also have the sides AC and BC and CF is the median on the third side, we get
\[ \Rightarrow AC + BC > 2CF{\text{ ......eq.(3)}}\]
Adding the equation (1), equation (2) and equation (3), we get
\[
   \Rightarrow AB + AC + AB + BC + AC + BC > 2AC + 2BE + 2CF \\
   \Rightarrow 2AB + 2AC + 2BC > 2AC + 2BE + 2CF \\
 \]
Taking 2 commons from L.H.S and R.H.S of the above equation, we get
\[ \Rightarrow 2\left( {AB + AC + BC} \right) > 2\left( {AC + BE + CF} \right)\]
Dividing the above equation by 2 into both sides, we get
\[
   \Rightarrow \dfrac{{2\left( {AB + AC + BC} \right)}}{2} > \dfrac{{2\left( {AC + BE + CF} \right)}}{2} \\
   \Rightarrow AB + AC + BC > AC + BE + CF \\
   \Rightarrow AB + BC + AC > AC + BE + CF \\
 \]

Hence, option B is correct.

Note:
First, draw the pictorial representation of the given problem for better understanding. You need to know the properties of triangles and their midpoint. Then we will use the properties accordingly. Then we will use the properties accordingly. This is a really simple problem, It is clear from the diagram that it is a triangle.