Question

In figure , AD is a median of a traingle ABC and AM per. to BC .Prove that :

AC² + AB² = 2AD²+ 1/2 BC²

Answer 1

$\implies Hey \:mate !$

BD = DC (AD is a median)

AC²= AM² +MC²

AC²= AM² +(CD+DM)²

AC²= AM² + CD² + DM² +2CD.DM .......(1)

In ∆ ABD ,

AB²=AM² + BM²

AB²=AM² + ( BD - MD )²

AB²=AM² + (BD² + MD² +2 BD.MD)

Add equations

AC² + AB² = AM² + AM² +MD²+MD² + DC² + BD² +2MD(DC-MD)

AC²+AB²= 2AM²+2MD²+DC²+DC²+2MD(DC-DC)

AC²+AB²=2(AM²+MD²)+1/2BC²+2MD(0)

$\boxed{AC^{2}+AB^{2}=2AD^{2}+\frac{1}{2}BC^{2} }$

Note [ AM²+MD²=AD² by pathagorus theorem]

[DC=BD = > DC²=BD²]

[DC+BD=BC => 2DC = BC due to median]

Hence proved

Answer 2

Heya!!

Your solution is in the attachments.

Hope it helps!!