X and Y are points on the sides AB and BC respectively of `triangleABC` such that XY||AC and XY divides `triangleABC` into two parts in area , find `(AX)/(AB)`

Explanation

Since XY||AC
In` triangleBXY and triangleBAC`.
`angle1=angle2` (corresponding angles)
`angleB= angleB` (common)
`triangleBXY~ triangleBAC` (AA corollary)
`(ar(triangleBXY))/(ar(triangleBAC))= (BX^(2))/(BA^(2))` (area ratio theorem) ....(1)
But `ar(triangleBXY)=1/2ar(triangleABC)` (given)
`Rightarrow" " (ar(triangleBXY))/(ar(triangleABC))=1/2`
from (1) and (2) we get
`(BX^(2))/(AB^(2))=1/2 Rightarrow (BX)/(AB) 1/(sqrt(2))`
` Rightarrow" "(AB-AX)/(AB)=1/sqrt2 Rightarrow (AB)/(AB)-(AX)/(AB)= 1/sqrt2`
` Rightarrow" " 1-(AX)/(AB)=1/sqrt2 Rightarrow(AX)/(AB)=1-1/sqrt2= (sqrt2-1)/sqrt2`
Hence , `(AX)/(AB)= (2-sqrt2)/2`