Question

if in a triangle ABC AD is median and AM|BC then prove AB^2+AC^2=2(AD^2+BD^2)

Answer 1

You know this is pretty basic geometry...

You have actually asked for the proof the famous Appolonius theorem...

You know what, just draw a perpendicular ( infact you already have one)

and apply pythagoras theorem and readjust to get the required.

We have,  AB^{2} = BM^{2} + AM^{2}  AC^{2} = CM^{2} + AM^{2}  adding together,   AB^{2} + AC^{2}= BM^2 + 2AM^2 + CM^2  AB^{2} + AC^{2}= (BD - MD)^2 + 2AM^2 + (CD+DM)^2  AB^{2} + AC^{2}=  2(AD^2+BD^2)

QED.  AB^{2} + AC^{2} = BD^2 + MD^2 - 2BD.MD + 2AM^2 + CD^2+DM^2 + 2CD.MD  AB^{2} + AC^{2} = BD^2 + MD^2  + 2AM^2 + CD^2 + DM^2  

AB^{2} + AC^{2} = 2BD^2 +(MD^2  + AM^2) + (DM^2 + AM^2)

AB^{2} + AC^{2} = 2BD^2 +AD^2 + AD^2