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AB//CD \(\Rightarrow S_{BAD}=S_{ABC}\Rightarrow S_{BAD}-S_{OAB}=S_{ABC}-S_{OAB}\Rightarrow S_{OAD}=S_{OBC}=10cm^2\)\(\dfrac{OB}{OD}=\dfrac{S_{OAB}}{S_{OAD}}=\dfrac{6}{10}=\dfrac{3}{5}\)

\(\dfrac{S_{OBC}}{S_{ODC}}=\dfrac{OB}{OD}=\dfrac{3}{5}\)

\(\Rightarrow S_{ODC}=\dfrac{5}{3}S_{OBC}=\dfrac{5}{3}.10=\dfrac{50}{3}\left(cm^2\right)\)

\(S_{ABCD}=S_{OAB}+S_{OAD}+S_{OBC}+S_{ODC}=6+10+10+\dfrac{50}{3}=\dfrac{128}{3}\left(cm^2\right)\)

\(\dfrac{S_{AIB}}{S_{AID}}=\dfrac{3}{9}=\dfrac{1}{3}\)

=>\(\dfrac{IB}{ID}=\dfrac{1}{3}\)

Vì ABCD là hình thang

nên \(S_{ADI}=S_{BCI}=9\left(cm^2\right)\)

Xét ΔIAB và ΔICD có

\(\widehat{IAB}=\widehat{ICD}\)(hai góc so le trong, AB//CD)

\(\widehat{AIB}=\widehat{CID}\)(hai góc đối đỉnh)

Do đó: ΔIAB~ΔICD

=>\(\dfrac{S_{IAB}}{S_{ICD}}=\left(\dfrac{IB}{ID}\right)^2=\dfrac{1}{9}\)

=>\(S_{ICD}=9\cdot3=27\left(cm^2\right)\)

Ta có: O nằm giữa B và D

=>\(\frac{S_{AOB}}{S_{AOD}}=\frac{OB}{OD}\)

=>\(\frac{OB}{OD}=\frac{6}{10}=\frac35\)

Vì AB//CD

nên \(\frac{OA}{OC}=\frac{OB}{OD}=\frac35\)

Vì \(\frac{OA}{OC}=\frac35\)

nên \(\frac{S_{BOA}}{S_{BOC}}=\frac35\)

=>\(\frac{6}{S_{BOC}}=\frac35\)

=>\(S_{BOC}=6\times\frac53=10\left(\operatorname{cm}^2\right)\)

Vì \(\frac{OA}{OC}=\frac35\)

nên \(\frac{S_{DOA}}{S_{DOC}}=\frac35\)

=>\(\frac{10}{S_{DOC}}=\frac35\)

=>\(S_{DOC}=10\times\frac53=\frac{50}{3}\left(cm^2\right)\)

\(S_{ABCD}=S_{OAB}+S_{OAD}+S_{OBC}+S_{OCD}\)

\(=6+10+10+\frac{50}{3}=26+\frac{50}{3}=\frac{78+50}{3}=\frac{128}{3}\left(\operatorname{cm}^2\right)\)

a:TAcó: O nằm giữa B và D

=>\(\frac{OB}{OD}=\frac{S_{AOB}}{S_{AOD}}=\frac{3.5}{10.5}=\frac13\)

OB/OD=1/3

=>\(\frac{S_{AOB}}{S_{BOC}}=\frac13\)

=>\(S_{BOC}=3\times S_{AOB}=10,5\left(\operatorname{cm}^2\right)\)

Vì \(\frac{OB}{OD}=\frac13\)

nên \(\frac{S_{BOC}}{S_{DOC}}=\frac13\)

=>\(S_{DOC}=10,5\times3=31,5\left(\operatorname{cm}^2\right)\)

\(S_{ABCD}=S_{OAB}+S_{OAD}+S_{OBC}+S_{OCD}\)

\(=3,5+10,5+10,5+31,5=14+42=56\left(\operatorname{cm}^2\right)\)

Vì AB//CD

nên \(\frac{OB}{OD}=\frac{AB}{CD}\)

=>\(\frac{AB}{CD}=\frac13\)

b: OM//DC

=>\(\frac{OM}{DC}=\frac{AO}{AC}\) (1)

ON//DC

=>\(\frac{ON}{DC}=\frac{BO}{BD}\) (2)

Vì AB//CD

nên \(\frac{OA}{OC}=\frac{OB}{OD}\)

=>\(\frac{AO}{AO+OC}=\frac{BO}{OB+OD}\)

=>\(\frac{AO}{AC}=\frac{BO}{BD}\) (3)

Từ (1),(2),(3) suy ra OM=ON

chịu thui

ko bt