The line segment XY is parallel to side AC of Δ ABC and it divides the triangle into two parts of equal areas. Find the ratio BXAB.
Given: XY || AC
Ar(△ABC)=2Ar(△XBY)
Here, ∠ BXY = ∠ A and ∠ BYX = ∠ C (Corresponding angles)
⇒ △ ABC ∼ △ XBY (AA similarity criterion)
⇒ Ar(△ABC)Ar(△XBY)=(ABXB)2
∵ Ar(△ABC)=2Ar(△XBY)
Ar(△ABC)Ar(△XBY)=21
⇒ (ABXB)2= 21
⇒ ABXB = √21
⇒XBAB=1√2
Rationalising the denominator we get,
XBAB=√22