Co ordinate Geometry Doubt Solve It!!

Arey yaar...to prove by second method requires stuff I did in 10th class...ab toh kuch yaad bhi nahi hai! damn.

Perhaps u could do it by concept of similsr triangles or mid pt. theorem.

1. Co ordinate Geometry

Let coordinates of B(0,0) , C (x,0), A(x₁,y₁).
Hence coordinates of mid points are D(x/2 , 0), E( (x+x₁)/2 , y₁/2 ), F( x₁/2 , y₁/2 ).

now BC²= (x-0)² + (0-0)² = x²
also EF² ={ (x+x₁)/2 - x₁/2}² + {y₁/2 - y₁/2}² = (x/2)² = x²/4

BC² = x² = 4
EF² x²/4

Hence proved.

2 . Euclidean Geometry (Normal Geometry)

Since D is mid point of BC
so BD = DC
BC = BD + DC = BD + BD = 2 BD
According to mid point theorem,
EF = BC/2 = 2 BD /2 = BD

BC² = (2 BD)² = 4 (BD)²
EF² = (BD)²

BC² = 4 (BD)² = 4
EF² (BD)²

Hence proved.

NO NO
HOW TO PROVE THAT
\frac{\Delta ABC}{\Delta DEF}=\frac{BC^2}{EF^2}

What you did was a piece of cake . just because pasted the whole sum does not mean that you solve it partially . PLEASE SOLVE IT