Question

Seg AN and seg CM are the medians of $\Delta$ ABC in which $\angle B={90}^o$. Prove that $4(A{N}^2+C{M}^2)=5A{C}^2$.

Answer 1

Proof: In right angled $\Delta ABC$
By Pythagoras theorem
$A{C}^2=A{B}^2+B{C}^2$ .......... (1)
In right angled $\Delta BMC$
By Pythagoras theorem
$C{M}^2=B{M}^2+B{C}^2$
$C{M}^2={\left(\frac{1}{2}AB\right)}^2+B{C}^2$ [M is the mid-point of AB]
$C{M}^2=\frac{1}{4}A{B}^2+B{C}^2$
$4C{M}^2=A{B}^2+4B{C}^2$ ......... (2)
(Multiplying by 4 on both sides)
In right angled $\Delta ABN$
By Pythagoras theorem
$A{N}^2=A{B}^2+B{N}^2$
$A{N}^2=A{B}^2+{\left(\frac{1}{2}BC\right)}^2$
$A{N}^2=A{B}^2+\frac{1}{4}B{C}^2$
$\therefore 4A{N}^2=4A{B}^2+B{C}^2$ ........... (3)
(Multiplying by 4 on both sides)
Adding (2) and (3)
$\therefore 4C{M}^2+4A{N}^2=A{B}^2+4B{C}^2+4A{B}^2+B{C}^2$
$\therefore 4(C{M}^2+A{N}^2)=5A{B}^2+5B{C}^2$
$\therefore 4(C{M}^2+A{N}^2)=5(A{B}^2+B{C}^2)$
$\therefore 4(C{M}^2+A{N}^2)=5A{C}^2$
$\therefore 4(A{N}^2+C{M}^2)=5A{C}^2$