View attachment 72418
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Dễ dàng chứng minh được: [tex]S_{AOD}=S_{BOC}[/tex]
Ta có: [tex]\frac{S_{AOD}}{S_{AOB}}=\frac{OD}{OB};\frac{S_{DOC}}{S_{COB}}=\frac{OD}{OB}\Rightarrow \frac{S_{AOD}}{S_{AOB}}= \frac{S_{DOC}}{S_{COB}}\Rightarrow S_{AOD}^2=S_{DOC}.S_{AOB}=a^2b^2\Rightarrow S_{AOD}=ab\Rightarrow S_{AOD}+S_{BOC}=2ab\Rightarrow S_{ABCD}=a^2+b^2+2ab=(a+b)^2[/tex]