Can Any One Give Me The Answer To My Question?

Question

In a parallelogram ABCD, E & F are the mid-points of sides AB&CD respectively. Show that the line segments AF & EC trisect the diagonal BD.

Answer 1

it's simple u can prove it by basic proportionality thm.

Answer 2

Consider AF & EC intersects BD at P & Q respectively. Consider triangle ABP & EBQ: AE=EB, EQ//AP. So, BQ=PQ. Similarly DP also will be equal to PQ. So, AF & EC trisect the diagonal BD.

Answer 3

FIRST WE MUST PROVE AF AND EC ARE PARALLEL.
WE DON'T KNOW IT WITHOUT PROVING IT.

Call the points on diagonal BD: N1 and N2.
N1 is AF intersected with BD
N2 is EC intersected with BD.
Extend AD and EC until they meet in say M.
Triangle MAE = Triangle ECB
(because AE=EB, the 2 E angles are equal
and angle C = angle M (because CB ll AD))
This shows that MA=CB
But CB=AD.
So MA=AD and CF=FD. So AFll EC.

So: ADN1 is similar (~) to DMN2.
Thus: DN1/N1N2=DA/AM=1.

So: DN1/N2N2=1.
This means that DN1=N1N2. (let's call this 1)

tRIANGLE EN2B is similar to triangle AN1B (because AF ll EC)
So: EB/EA=BN2/N2N1. So: N2N1=BN2. (call this 2)

From 1 and 2 we have DN1=N1N2=N2B

Answer 4

I am certain that it is true. I was able to prove it with a rectangle, but it's going to take too long to do it with a parallelogram.

Basically I did it by forming lines, and then proving that the lines intersect at the right distance from D / E.

Answer 5

oh cmon, refer to the math books for all these stupid theorems.. they never made sense anyways..

Answer 6

Good luck with that!