In ΔABC, XY || AC and XY divides the triangle into two parts of equal area. Find the ratio of AX/XB.
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Given :
In ∆ABC , XY // AC and XY divides
∆ABC into two parts of equal area .
AX/XB = ?
proof :
In ∆XBY , ∆ABC ,
<B is common angle ---( 1 )
BC/BA = BY/BC ----( 2 )
[ By Thales theorem ]
So, ∆XBY ~ ∆ABC
( SAS similarity )
(Area ∆XBY)/(area ∆ABC )=BX²/AB²
[area ∆XBY = (area ∆ABC )/2---( 3 )]
(area ∆ABC/2)/(area ∆ABC ) =BX²/AB²
=> 1/2 = BX²/AB²
BC/BA = √(1/2)
BC/AB = 1/√2
=> AB/XB = √2
[ But , AB = AX + XB ]
=> ( AX+XB )/XB = √2
AX/XB + XB/XB = √2
AX/XB + 1 = √2
AX/XB = √2 - 1
Therefore ,
AX /XB = ( √2 - 1 )/1
I hope this helps you.
: )
In ∆ABC , XY // AC and XY divides
∆ABC into two parts of equal area .
AX/XB = ?
proof :
In ∆XBY , ∆ABC ,
<B is common angle ---( 1 )
BC/BA = BY/BC ----( 2 )
[ By Thales theorem ]
So, ∆XBY ~ ∆ABC
( SAS similarity )
(Area ∆XBY)/(area ∆ABC )=BX²/AB²
[area ∆XBY = (area ∆ABC )/2---( 3 )]
(area ∆ABC/2)/(area ∆ABC ) =BX²/AB²
=> 1/2 = BX²/AB²
BC/BA = √(1/2)
BC/AB = 1/√2
=> AB/XB = √2
[ But , AB = AX + XB ]
=> ( AX+XB )/XB = √2
AX/XB + XB/XB = √2
AX/XB + 1 = √2
AX/XB = √2 - 1
Therefore ,
AX /XB = ( √2 - 1 )/1
I hope this helps you.
: )
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