through the midpoint m of the side cd of a paralelogram abcd,the line bm is drawn intersecting ac in l and ad produced in e.prove that el=2bl

thanks
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Thanx rellly nice answer quite helpful
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What is a^??? In the ans been given ?
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Gud
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Gud
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thank u very much sister
 
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How you can say that BC=DE. Please explain
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ΔBCM & ΔEDM are similar.        MD=CM  =>  BC = DE,   as AD = BC,  AE = 2 BC
ΔACL & ΔAEL are similar.        EL / BL = AE / BC = 2
So     EL = 2 BL

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Thanks it was helpful
 
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Amit, MD=CO, because they are the sides opposite to equal angles.. Angle BMC = Angle DME [VERTICALLY OPPOSITE ANGLES]
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Guuh
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👍👍👍👍👍hope this will help....

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Thanks
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This is done in the simplest way, hope all will understand
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How can ACL be a triangle???
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Hope it is helpful Cheers✌️✌️

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Given: ABCD is a parallelogram where M is the midpoint of side CD. BM is drawn intersecting diagonal AC in Land AD produced in E. To prove: EL = 2BL Proof: In ∆DME and ∆CMB EDM = BCM (pair of alternate angles) DM = CM (M is the midpoint of CD) DME = BMC (vertically opposite angles) ∴∆DME ≅ ∆CMB (ASA congruence criterion) ⇒ DE = BC (c. p. c. t) Now, In ∆ALE and ∆BLC, ALE = BLC (vertically opposite angles) AEL = LBC (pair of alternate angles) ∴∆ALE ∼ ∆CLB (AA similarity criterion) Al/cl=el/bl=ae/cb El/bl=ae/bc El/bl=ad+de/bc El/bl=2bc/bc El=2bl
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How acl can be a triangle
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Shashank
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This answer was really helpful for me..... thanks a lot..
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Bhosadpapu khinka l**u
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same as neha
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Okkk
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Answer is below

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This the real answer for this question friends.

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Hope it will help you.. Thumbs up plzz

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