ABCD is a rectangle. From 'C' two lines are drawn to meet AB and AD at E and F respectively (here AB and AD are not produced). From B, a line is drawn to meet CE, DE and AD at G, H, and F respectively. Another line which passes through D and H meets CF and AB at J and E respectively. If the area of triangle BEG is 500, area of triangle JFD is 405 and the area of quadrilateral EHFA is 1110, then find the area of quadrilateral CJHG.
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$\begingroup$ You've been here long enough <g>: What have you tried? $\endgroup$ – Brian Tung Aug 22 '15 at 17:23
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$\begingroup$ tried a bit not getting the idea to solve $\endgroup$ – lokesh sangabattula Aug 22 '15 at 17:25
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Let the area of the figures be
$$[AFHE]=a,[EHG]=b,[EGB]=c,$$ $$[FHJ]=d,[HJCG]=e,[GCB]=f,$$ $$[FDJ]=g,[DJC]=h.$$
From $[DEC]=[ADE]+[EBC]$, $$b+e+h=a+d+g+c+f\tag1$$
From $[FBC]=[AFB]+[DFC]$, $$d+e+f=a+b+c+g+h+c\tag2$$
Now $(1)+(2)$ gives $$b+e+h+d+e+f=a+d+g+c+f+a+b+c+g+h+c,$$ i.e. $$[HJCH]=e=a+g+c=500+405+1110=\color{red}{2015}.$$