D, E,F are the midpoints of sides BC, CA and AB of Triangle ABC. Show that BE anf CF bisects DF, and DE resp

Answer 1

EF is joined
F and E are the midpoints of Sides AB and AC respectibely
Therefore FE=1/2 BC and ||BC
in quadrilateral BDEF FE=1/2BC=BD and||BD
Therefore BDEF is a parallelogram and DF and BE are its diagonals.
We know that diagonals of a parallelogram bisects each other
therefore BE bisects DF
Similarly it can be proved that CF bisects DE

Answer 2

Let the intersection of CF and ED be G.
Then the triangle CGE similar triangle CFA.
Therefore CE/CA = CG/ CF
But CE=1/2CA since E is midpoint of CA.
Therefore 1/2CA/CA =1/2 --> CG=CF
Thus EG = 1/2AF and GD = 1/2FB
But AF = FB
Thus EG = GD
Therefore CF bisects ED
In like manner it can be shown that BE bisects DF.
Q.E.D

Answer 3

∆BDF is similar to ∆BCA. Since E is the midpoint of CA, BE bisects CA, and must therefore bisect DF.. The same proof applies to lines CF and DE

Answer 4

It is difficult to explain a geometric problem on the internet, but I will give you some tips:

1. Use mid point triangle theorem

2. Find the congruent triangles

3. deduce that BE & CF bisect DE & DF

If it sounds too complicated, email me.

Answer 5

wrong, when done possible vise versa, you get the same awnser, but if you look carefully you will see what i mean
!

Answer 6

you may flunk in your maths exams.....get hold of a good tution master..