3
$\begingroup$

In a parallelogram ABCD. M is the midpoint of CD. Line BM intersects AC at L and it also intersects AD extended at E. Prove that EL=2BL

PS: This is not a homework problem. I was solving geometry for fun. I'm unable to solve this. :(

  • 0
    Can someone think of a better title?2010-11-10
  • 0
    No! but if you can please comment.2010-11-11
  • 0
    I couldn't either! Hence the request. Maybe something like "Line joining midpoint of other side". EDIT: changed.2010-11-11

3 Answers 3

1

Given:

  1. $ABCD$ is a parallelogram

  2. $M$ is the midpoint of $CD$.

To prove that:

  • $EL = 2BL$

Proof:

  • In $ΔDME$ and $ΔCMB$

$

$DM=CM$ (midpoint of M)

$

Therefore $ΔDME$ congruent to $ΔCMB$ (ASA congruence)

$DE=BC$(c.p.c.t)

In $ΔALE$ and $ΔBLC$

$

$

Therefore $ΔALE$ congruent to $ΔBLC$ (AA congruence)

$AE/BC=AL/LC=EL/BL$ (corresponding sides)

$EL/BL=AE/FC$

$EL/BL=AD+DE/BC$

$EL/BL=BC+BC/BC$ ($AD=BC$ and $BC=DE$)

$EL/BL=2BC/BC$

$EL/BL=2$

$EL=2BL$

Hence proved.

4

Let BD and AC intersects at point O. We have triangle BCD. BO=OD because ABCD is parallelogram. In this triangle CO and BM are medians. Medians intersects in one point and divide each other in relation 2:1. So BL = 2LM.

Also BM = ME thus triangle MBC = triangle MDE because they are congruent and DM = CM.

So BL = 2/3 BM => BL = 2EL.

3

Here is a different way to look at it.

Make two copies of the parallelogram, sharing the side CD.

alt text

The second copy is A'B'DC.

We will show that B'=E.

Consider $\triangle{B'AB}$ and $\triangle{B'DM}$. These are similar and since B'D = AD, we have that DM = AB/2. M is the midpoint of CD and thus B' = E.

Thus B'L' = BL.

Now consider $\triangle{B'DL'}$ and $\triangle{B'AL}$. These are similar and so B'L' = LL', as B'D = AD.

Thus EL = EL' + LL' = 2B'L' = 2BL.