In \(\Delta ABC\), A – X – B and A – Y – C with seg XY || side BC. If \(A(\Delta AXY) : A(\square XBCY) = 1 : 8\), find AX : XB.

Answer: AX : XB = 1 : 2

XY || BC, so \(\angle AXY \cong \angle ABC\) (corresponding angles) and \(\angle A\) is common. \(\therefore \Delta AXY \sim \Delta ABC\) (AA test).

Let \(A(\Delta AXY) = k\); then \(A(\square XBCY) = 8k\) and \(A(\Delta ABC) = k + 8k = 9k\).

\(\dfrac{A(\Delta AXY)}{A(\Delta ABC)} = \dfrac{AX^2}{AB^2}\) \(\therefore \dfrac{k}{9k} = \dfrac{AX^2}{AB^2}\) \(\therefore \dfrac{AX}{AB} = \dfrac{1}{3}\).

Let AX = t, AB = 3t; then XB = 3t − t = 2t. \(\therefore AX : XB = 1 : 2\).

Tip: always change the trapezium’s area into the whole triangle’s area before using the squares rule.