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2015-11-13T11:35:36+05:30
since M is the mid point of AB , AM = MB
AB = 2 AM = 2 MB
since ABX and ABY are equilateral , AB = BX = AX = 2MY = 2BY
triangles BXZ and BZY have congruent vertical angles at Z , and congruent 60
° angles ZBX and ZMY .
they are similar , and since BX = 2BY ,ZB = 2MZ
then AM = MB = MZ +ZB = 3 MZ 
so, AZ = AM +MZ = 3MZ+MZ = 4MZ ,
and since ZB = 2MZ ,AZ/ZB = 2MZ / 2MZ = 2 
AZ / ZB = 2 
AZ = 2ZB
therefore proved.
  i hope it helps u ...................^_^

2 5 2
but i can't edit it now sir 
in general, it will be helpful to do so
ok sir from next time i will do it 
Thnx it really helped me
ur welcome frnd :)
2015-11-13T15:29:05+05:30

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See diagram.

Use similar triangles principles to solve this.  Let  AB = a.  AM = MB = a/2

Method 1:

Draw a straight line XMW such that  MW is perpendicular to YW.
MX = √3 a/2 in triangle AXB.
MW is parallel  to the altitude of triangle YMB, from Y on to MB.  MW = √3 a / 4
XW = MX + MW = 3√3 a/4

WY is parallel to MB and is  =  MB / 2

So from the two similar triangles  ΔXMZ and Δ XWY:

     MZ / MX = WY / WX   
     MZ = MX * WY / WX = (√3a/2) * (a/4) / (3√3 a/4)
           = a/6

So  AZ =  a/2 + a/6 = 2 a/3
 and  ZB = a/2 - a/6 =  a/3

Hence ,  AZ = 2 * ZB

Method 2:

 You could take the similar triangles  MZY and  BZX.  to prove  BZ = 2 * MZ
         (as  MY = a/2  and  BX = a).
   Then the result follows.

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