# Help me

1. Nov 7, 2005

### vaishakh

Please help me with this question. ABC is a triangle. D, E and F are points on BC, AC and AB such that AD, BE and CF are concurrent. Show that lines parallel to AD, BE and CF from the midpoints BC, AC and AB are also concurrent.
What I have been taught is the Stewart's theorem (for three collinear points A, B and C and any point P, AP^2*BC + BP^2*CA + CP^2*AB AB*BC*CA = 0), Meanaleuas theorem (a line joining three sides of a triangle divides it in such a way that the product of the ratio of their division is 1 and the sign is negative in vector notations) and Cava's theorem (the lines joining the vertices of a triangle meets the opposite side in such a way that the product of the ratio is 1). Wherever ratios are mentioned they are taken in the cyclic form like-AD/DB * BE/EC * AF/FC and not in the way - AD/BD * BE/CE * AF/CF. I am also expected to know the standard rules of parallel lines and intercepts while solving the question.
Can you help me from this knowledge?

(i am sorry if moderators feel homework should not be submitted here)

2. Nov 14, 2005

### vaishakh

I think no one is helping me because I have not cleared what I did. There is nothing much to be cleared. The question is of a type where I don't know how to start with the question. Even the fact that I have drawn a diagram to that question itself is a considerable part of my attempt.
If D', E' and F' are the midpoints of BC, AC and AB respectively, then if line D' and E' intersect at P', then that F'P' is parallel to CF is what I need to prove. The point of concurrence of AD, BC and AC is taken to be P. I tried to find the relation between the parallelogram formed as also tried to locate any similar triangles without any success