Question

The sum of the squares of two sides of a triangle is equal to twice the square on half the third side plus twice the square on the median which bisects the third side (Appolonius theorem).

Answer 1

R.E.F image

Theorem:- The sum of the square of two of a triangle

is equal to twice the square on hall the third side plus

twice the square on the median which bisects the

third side.

Proof:- Given $△ABC$ and $AD$ is a median

We need to proof $A{B}^2+A{C}^2=2A{D}^2+2{\left(\frac{1}{2}BC\right)}^2$

i.e, $A{B}^2+A{C}^2=2A{D}^2+2B{D}^2$.........(i)

Let $AN\perp BC$

Then $ln△ABN,A{B}^2=A{N}^2+B{N}^2$

ln $△ANC,A{C}^2=A{N}^2+N{C}^2$...........(ii)

Add (i) and (ii)

We get $A{B}^2+A{C}^2=A{N}^2+B{N}^2+A{N}^2+N{C}^2$

$=2A{N}^2+B{N}^2+(DC-DN{)}^2$

$=2A{N}^2+(BD+DN{)}^2+(DC-DN{)}^2$

$=2A{N}^2+B{D}^2+D{N}^2+2.BD.DN+D{C}^2+D{N}^2-2DC.DN$

$=2A{N}^2+2D{N}^2+B{D}^2+D{C}^2-2DC.DN+2BD.DN$

$=2(A{N}^2+D{N}^2)+B{D}^2+B{D}^2-2DC.DN+2BD.DN$

$=[\because BD=DC]$

$\therefore A{B}^2+A{C}^2=2A{D}^2+2B{D}^2$

$A{B}^2+A{C}^2=2A{D}^2+2{\left(\frac{1}{2}BC\right)}^2$