Question

In ΔABC, A-X-B and A-Y-C are such that segment XY || side BC.

Segment XY divides Δ ABC into two equal areas, then find `"BX"/"AB"`.

Answer 1

Area (ΔABC) = 2 Area (ΔAXY)

`(Area (ΔAXY))/(Area (ΔABC)) = 1/2`    ...(Equation 1)

Now, in ΔAXY and ΔABC,

∠A = ∠A    ...[Common]

∠AXY = ∠ABC    ...[Corresponding Angles]

∴ AA similarity condition

ΔAXY ∼ ΔABC

`(Area (ΔAXY))/(Area (ΔABC)) = (AX^2)/(AB^2)`    ...(Equation 2)

Comparing Equation 1 with Equation 2:

`(AX^2)/(AB^2) = 1/2`

`((AX)/(AB))^2 = 1/2`

`(AX)/(AB) = 1/sqrt2`

`(AB - BX)/(AB) = 1/sqrt2`

`(AB)/(AB) - (BX)/(AB) = 1/sqrt2`

`1 - (BX)/(AB) = 1/sqrt2`

`(BX)/(AB) = 1 - 1/sqrt2`

∴ `(BX)/(AB) = (sqrt2 - 1)/sqrt2`