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In parallelogram ABCD point K belongs to diagonal BD so that BK:DK=1:4. If the extension of AK meets BC at point E, what is the ratio of BE:EC?

 Dec 25, 2018
 #1
avatar+94453 
+2

Let ABCD be the parallelogram

 

Angle BKE = Angle AKD   (vertical angles)

Since AB is parallel to CD  and BD is a transversal cuting these parallels, then

Angle KBE = Angle ADK

 

Then ΔAKD  is similar to ΔEKB

 

And  

 

BK / DK  =  BE / AD

 

But since we have a paralleogram, AD = BC

 

So

 

BK / DK =  BE / BC

 

But  BK /DK  =  1/4

 

So

 

BE / BC   =  1/4

 

So

 

BC / BE  =  4/1

 

[ BE + EC ]  / BE =   4

 

BE / BE  +  EC / BE  = 4

 

1 + EC / BE  =  4

 

EC / BE =  3

 

EC / BE = 3/1

 

So

 

BE / CE   =   1 / 3

 

 

cool cool cool

 Dec 26, 2018

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