Пусть СН -высота в [tex]\triangle[/tex]DOC и AN -высота в [tex]\triangle[/tex]AOD .
[tex]S_{ABEO }[/tex]=S=?
[tex]\frac{ S_{EOC } }{ S_{DOC } }= \frac{2}{3}[/tex] ; [tex]\frac{ \frac{OE.CH}{2} }{ \frac{DO.CH}{2} } =\frac{2}{3}[/tex]
[tex]\frac{OE}{DO} = \frac{2}{3}[/tex]
(1) Мы знаем [tex]S_{ABC }= S_{ACD }[/tex]
S+2 =[tex]S_{AOD }[/tex]+3 ;[tex]S_{AOD }[/tex]= S-1
(2)[tex]S_{AEO }[/tex]= ?
[tex]\frac{ S_{AEO } }{ S_{AOD } } =\frac{ \frac{OE.AN}{2} }{ \frac{DO.AN}{2} }[/tex]=[tex]\frac{OE}{DO} =\frac{2}{3}[/tex]
[tex]\frac{ S_{AEO } }{S-1}= \frac{2}{3}[/tex] ; [tex]S_{AEO } = \frac{2S-2}{3}[/tex]
(3) [tex]S_{AED } = \frac{ S_{ABCD } }{2}[/tex]
[tex]S_{AEO }+ S_{AOD }= \frac{2+3+S+(S-1)}{2}[/tex] ; [tex]\frac{2S-2}{3} +S-1 = \frac{2S+4}{2}[/tex] ; [tex]\frac{2S-2}{3}[/tex]=3
S=[tex]\frac{11}{2}[/tex]
Мы можем найти [tex]S_{ABCD }[/tex]
[tex]S_{ABCD }[/tex]=2+3+S+(S-1) = 5+[tex]\frac{11}{2} + \frac{11}{2}[/tex]-1 =4+11
[tex]S_{ABCD }[/tex] =15