1Perhaps u could do it by concept of similsr triangles or mid pt. theorem.
Prove it from both Sides : 1. Co ordinate Geometry
2 . Euclidean Geometry (Normal Geometry)
Separately
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7 Answers
39Arey yaar...to prove by second method requires stuff I did in 10th class...ab toh kuch yaad bhi nahi hai! damn.
1
11. 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.
12 . 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.
1Thanks Muskaan
I hate you man -- Pritish Go Study in class X
JUST KIDDIN
1NO 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
62the last part isnt very difficult either..
once you have the coordinates.. it doesnt take a lot except some calculations that the area ratios will be same as the square of the distances...
just apply the formula for area given three points of a triangle...