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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. :(

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Can someone think of a better title? – Aryabhata Nov 10 '10 at 22:56
No! but if you can please comment. – claws Nov 11 '10 at 7:51
I couldn't either! Hence the request. Maybe something like "Line joining midpoint of other side". EDIT: changed. – Aryabhata Nov 11 '10 at 15:12
up vote 3 down vote accepted

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.

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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.

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  1. $ABCD$ is a parallelogram

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

To prove that:

  • $EL = 2BL$


  • In $ΔDME$ and $ΔCMB$

$<EDM=<BCM$ (alt.angles)

$DM=CM$ (midpoint of M)

$<DME=<BMC$ (V.O.A)

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


In $ΔALE$ and $ΔBLC$

$<ALE=<BLC$ (V.O.A)

$<AEL=<LBC$ (alt. angles)

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

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



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




Hence proved.

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