In ΔABC, DE || BC. AD = x, DB = x − 2, AE = x + 2 and EC = x − 1. Find the value of x.
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AD/BD=AE/EC
=X/X-2=X+2/X-1
=X(X-1)=(X-2)(X+2)
=X²-X=X²-4
X²-X²=-4+X
-4+X=0
X=4
Answered by
14
Solution :
Given :
In ∆ABC , DE//BC
AD = x , DB = x - 2 ,
AE = x + 2 , EC = x - 1 ;
In ∆ABC ,DE//BC
then
AE/EC = AD/DB
[ Thales theorem ]
( x + 2 )/( x - 1 ) = x/( x - 2 )
=> ( x + 2 )( x - 2 ) = x( x - 1 )
=> x² - 4 = x² - x
=> -4 = - x
Therefore ,
x = 4
Given :
In ∆ABC , DE//BC
AD = x , DB = x - 2 ,
AE = x + 2 , EC = x - 1 ;
In ∆ABC ,DE//BC
then
AE/EC = AD/DB
[ Thales theorem ]
( x + 2 )/( x - 1 ) = x/( x - 2 )
=> ( x + 2 )( x - 2 ) = x( x - 1 )
=> x² - 4 = x² - x
=> -4 = - x
Therefore ,
x = 4
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