Question

A(x₁,y₁), B(x₂,y₂), C(x₃,y₃) are the vertices of ΔABC and D, E, F are the mid-points of the sides BC, CA, AB respectively. Prove that ؈DEFB is a parallelogram.

Answer 1

F is the midpoint of AB ,
from midpoint section formula,
F = {(x1 + x2)/2 , (y1 +y2)/2}

similarly, E is the midpoint of AC,
so, E = {(x1 + x3)/2, (y1 + y3)/2}

and D is the midpoint of BC.
so , D = {(x2 + x3)/2, (y2 + y3)/2}

now, we have []DEFB where , D = {(x2+x3)/2,(y2+y3)/2} , E = {(x1+x3)/2,(y1+y3)/2} ,F = {(x1+x2)/2, (y1+y2)/2} and B = (x2,y2)

we know, diagonals of parallelogram bisect each other at a point.
in simple way : in Parallelogram, midpoint of 1st diagonal should be equal midpoint of 2nd diagonal .

midpoint of diagonal DF = [{(x2+x3)/2 + (x1+x2)/2}/2, {(y2+y3)/2 + (y1+y2)/2}/2]
= {(x1 + 2x2 + x3)/4, (y1 + 2y2 + y3)/4}

midpoint of diagonal EB = [{(x1+x3)/2+x2}/2 , {(y1+y3)/2+y2}/2]
= {(x1+2x2+x3)/4 , (y1+2y2+y3)/4}

here it is clear that midpoint of DF = midpoint of EB
so, []DEFB is parallelogram.

Answer 2

From the attachment above ,

We know that ,

( 1 ) = ( 2 )

Therefore ,

In a DEFB Quadrilateral ,

Diagonals BE , DF are bisects each

other .

DEFB is a parallelogram .

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