Question 1188243
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Draw a line perpendicular to BD from E.  Call this h.

Area of parallelogram is |CD|*h = 2h
Area of triangle BDE = (1/2)|BD|*h = 3h

Triangle BDE is similar to triangle BCA.
(Area triangle BCA) = {{{k^2}}} * (Area triangle BDE)
where  k = |BC|/|BD| = (6+2)/6 = 8/6 = 4/3
thus  {{{k^2}}} = 16/9  

Putting it all together:   
(Area parallelogram)/(Area triangle BCA) = 2h / (3h * 16/9) = 3/8