From the diagram, XYZQ is a parallelogram. Thus, |YZ| = |XQ| = |PX|; \(\bigtriangleup\)PXY, Let the area of XYZQ = A₁, the area of \(\bigtriangleup\)PXY
= Area of \(\bigtriangleup\)YZR = A₂
Area of \(\bigtriangleup\)PQR = A = A₁ + 2A₂
But from similarity of triangles
\(\frac{\text{Area of PQR}}{\text{Area of PXY}} = (\frac{PQ}{PX})^2 = (\frac{QR}{XY})^2\)
\(\frac{A}{A_2} = (\frac{2}{1})^2 = \frac{2}{1}\)
A = 4A₂ But, A = A₁ + 2A_{[/sub]
A[sub]1} = 4A₂ - 2A₂
A₁ = 2A₂
\(\frac{A_1}{A_2}\) = 2
A₁:A₂ = 2:1
Area of XYZQ:Area of \(\bigtriangleup\)YZR = 2:1