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Solve:- 
Q. In the following triangle ABC, BN is the median.
Derive that the median of the triangle is:
$\frac{1}{2}\sqrt{2a^2+2c^2-b^2}$

 
Hi, 
This question needs the use of appolonius theorem so lets prove that first and then solve this sum, ;-
In any triangle the sum of the squares of any two sides is equal to twice the square of half the third side together with twice the square of the median.

given:  ABC is a triangle and AD is the median of the triangle.

TPT : 

construction: draw 

proof:

in the right angled triangle ABE :

 ............(1)

in the right angled triangle ACE:

 ..........(2)

in the right angled triangle AED:

...........(3)

adding eq(1) and eq(2):

DC = BD

which is the required result.


So now here in our question, 
$$\begin{gathered} {\mathrm{BC}}^2+{\mathrm{BA}}^2= 2\left({\mathrm{BN}}^2+{\mathrm{AN}}^2\right) \\ {\mathrm{a}}^2+{\mathrm{c}}^2= 2\left({\mathrm{m}}^2+{\left(\frac{\mathrm{b}}{2}\right)}^2\right) \left[\mathrm{Since} \mathrm{CN}\hspace{0.17em}= \mathrm{NA}\hspace{0.17em}= \frac{\mathrm{AC}}{2}\right] \\ {\mathrm{a}}^2+{\mathrm{c}}^2=2{\mathrm{m}}^2+\frac{{\mathrm{b}}^2}{2} \\ 2{\mathrm{m}}^2={\mathrm{a}}^2+{\mathrm{c}}^2-\frac{{\mathrm{b}}^2}{2}=\frac{2{\mathrm{a}}^2+2{\mathrm{c}}^2-{\mathrm{b}}^2}{2} \\ {\mathrm{m}}^2=\frac{2{\mathrm{a}}^2+2{\mathrm{c}}^2-{\mathrm{b}}^2}{4} \\ \mathrm{m} = \frac{1}{2}\sqrt{2{\mathrm{a}}^2+2{\mathrm{c}}^2-{\mathrm{b}}^2} \end{gathered}$$