Let ABCD be a convex quadrilateral. Let P and Q be points on side \(\overline{AB}\) such that AP = PQ = QB. Similarly, BR = RS = SC, CT = TU = UD, and DV = VW = WA.
The area of quadrilateral ABCD is 180. Find the area of hexagon AWTCSP.
+23246 Draw line segment AC, dividing the figure into two triangles.
In the top triangle, triangle(BPS) ~ triangle(BAC) [side-angle-side for similarity]
Since BP = (2/3)rds BA, the area of triangle(BPS) = (2/3 · 2/3)·triangle(BAC) = 4/9·triangle(BAC)
Similarly, the area of triangle(WDT) = 4/9·triangle(ADC).
Adding these two parts together, we get the fact that the unshaded area is 4/9ths the quadrilateral.
However, we want the shaded area, so it will be 5/9ths the quadrilateral.
(5/9)·180 = 100
