Lấy M là trung điểm của DB

=>AD=DM=MB=1/3AB

Xét ΔAMC có AD/AM=AE/AC

nên ΔADE đồng dạng với ΔAMC

=>\(\dfrac{S_{ADE}}{S_{AMC}}=\left(\dfrac{AE}{AC}\right)^2=\dfrac{1}{4}\)

=>\(S_{AMC}=40\left(cm^2\right)\)

AM=2/3AB

=>\(S_{ABC}=\dfrac{3}{2}\cdot S_{AMC}=60\left(cm^2\right)\)

tự đi mà làm :))

a: AE+EC=AC

=>\(EC+\dfrac{2}{5}AC=AC\)

=>\(EC=\dfrac{3}{5}AC\)
\(\dfrac{AE}{EC}=\dfrac{\dfrac{2}{5}AC}{\dfrac{3}{5}AC}=\dfrac{2}{5}:\dfrac{3}{5}=\dfrac{2}{3}\)

Xét ΔACB có IE//AB

nên \(\dfrac{IC}{IB}=\dfrac{EC}{EA}=\dfrac{3}{2}\)

b: Xét ΔACB có IE//AB

nên \(\dfrac{IE}{AB}=\dfrac{CI}{CB}=\dfrac{3}{5}\)

AD+DB=AB

=>\(DB+\dfrac{2}{3}AB=AB\)

=>\(DB=\dfrac{1}{3}AB\)

=>AB=3BD

\(\dfrac{IE}{AB}=\dfrac{3}{5}\)

=>\(\dfrac{IE}{3BD}=\dfrac{3}{5}\)

=>\(\dfrac{IE}{BD}=\dfrac{9}{5}\)

Xét ΔFEI có DB//EI

nên \(\dfrac{FD}{FE}=\dfrac{DB}{EI}=\dfrac{5}{9}\)

=>\(FD=\dfrac{5}{9}FE\)

FD+DE=FE

=>\(DE+\dfrac{5}{9}FE=FE\)

=>\(DE=\dfrac{4}{9}FE\)

\(\dfrac{DF}{DE}=\dfrac{\dfrac{5}{9}EF}{\dfrac{4}{9}EF}=\dfrac{5}{9}:\dfrac{4}{9}=\dfrac{5}{4}\)

c: CI/IB=3/2

=>CI=3/2BI

BI+CI=BC

=>\(BC=\dfrac{3}{2}BI+BI=\dfrac{5}{2}BI\)

Xét ΔFEI có DB//EI

nên \(\dfrac{FB}{BI}=\dfrac{FD}{DE}=\dfrac{5}{4}\)

=>\(FB=\dfrac{5}{4}BI\)

mà \(BC=\dfrac{5}{2}BI\)

nên \(\dfrac{FB}{BC}=\dfrac{\dfrac{5}{4}BI}{\dfrac{5}{2}BI}=\dfrac{5}{4}:\dfrac{5}{2}=\dfrac{1}{2}\)

=>\(\dfrac{FB}{FC}=\dfrac{1}{2+1}=\dfrac{1}{3}\)

_{vi S deb=Sdec [chung đáy DE;chiều cao hạ từ đỉnh B hoặc C xuống đáy DE nên bằng nhau]}

_{.Vì S deb và Sdec đều chung tam giác DEI nên S DIB =  S EIC}

và câu hỏi là gì vậy bạn ?

Ta có: CE+EB=CB

=>\(EB=BC-\frac23\times BC=\frac13\times BC\)

=>CE=2xEB

=>\(EB=\frac12\times EC\)

=>\(S_{AEB}=\frac12\times S_{AEC};S_{OEB}=\frac12\times S_{OEC}\)

=>\(S_{AEB}-S_{OEB}=\frac12\times\left(S_{AEC}-S_{OEC}\right)\)

=>\(S_{AOB}=\frac12\times S_{AOC}\) (1)

DA+DB=AB

=>\(DB=AB-\frac13\times AB=\frac23\times AB\)

=>\(DB=2\times DA\)

=>\(S_{CDB}=2\times S_{CDA};S_{ODB}=2\times S_{ODA}\)

=>\(S_{CDB}-S_{ODB}=2\times\left(S_{CDA}-S_{ODA}\right)\)

=>\(S_{COB}=2\times S_{AOC}\) (2)

Từ (1),(2) suy ra \(\frac{S_{BOA}}{S_{BOC}}=\frac12:2=\frac14\)

=>\(S_{BOA}=\frac14\times S_{BOC}\)

Vì \(EB=\frac13\times BC\)

nên \(S_{OEB}=\frac13\times S_{BOC}\)

=>\(\frac{S_{EBO}}{S_{BOA}}=\frac13:\frac14=\frac43\)

=>\(\frac{OE}{OA}=\frac43\)

=>\(OE=\frac43\times OA\)

Ta có: OA+OE=AE

=>\(AE=OA+\frac43\times OA=\frac73\times OA\)

=>\(AO=\frac37\times AE\)

=>\(S_{ADO}=\frac37\times S_{ADE}\)

Vì \(AD=\frac13\times AB\) nên \(S_{ADE}=\frac13\times S_{AEB}\)

=>\(S_{ADO}=\frac37\times\frac13\times S_{AEB}=\frac17\times S_{AEB}\)

Vì BE=1/3BC

nên \(S_{AEB}=\frac13\times S_{ABC}\)

=>\(S_{ADO}=\frac17\times\frac13\times S_{ABC}=\frac{1}{21}\times S_{ABC}\)

Vì \(CE=\frac23\times CB\)

nên \(S_{CEO}=\frac23\times S_{COB}\)

Ta có: \(S_{AOB}+S_{AOC}+S_{BOC}=S_{ABC}\)

=>\(S_{ABC}=S_{AOC}+2\times S_{AOC}+\frac12\times S_{AOC}=\frac72\times S_{AOC}\)

=>\(\frac{S_{COB}}{S_{ABC}}=2:\frac72=\frac47\)

=>\(S_{BOC}=\frac47\times S_{ABC}\)

=>\(S_{CEO}=\frac23\times\frac47\times S_{ABC}=\frac{8}{21}\times S_{ABC}\)

=>\(\frac{S_{ADO}}{S_{CEO}}=\frac{1}{21}:\frac{8}{21}=\frac18\)

Ta có: \(AD+DB=AB\)

=>\(DB=AB-AD=AB-\frac13\times AB=\frac23\times BA\)

Ta có: BE+EC=BC

=>\(EC=BC-BE=BC-\frac13\times BC=\frac23\times BC\)

Ta có: CG+GA=CA

=>\(GA=CA-CG=CA-\frac13\times CA=\frac23\times CA\)

Ta có: \(BE=\frac13\times BC\)

=>\(S_{AEB}=\frac13\times S_{ABC}\)

Ta có: \(BD=\frac23\times BA\)

=>\(S_{BDE}=\frac23\times S_{AEB}=\frac23\times\frac13\times S_{ABC}=\frac29\times S_{ABC}\)

Ta có: \(CE=\frac23\times CB\)

=>\(S_{AEC}=\frac23\times S_{ACB}\)

Ta có: \(CG=\frac13\times CA\)

=>\(S_{EGC}=\frac13\times S_{AEC}=\frac13\times\frac23\times S_{ABC}=\frac29\times S_{ABC}\)

Ta có: \(AD=\frac13\times AB\)

=>\(S_{ADC}=\frac13\times S_{ABC}\)

Vì \(AG=\frac23\times AC\)

nên \(S_{AGD}=\frac23\times S_{ADC}=\frac23\times\frac13\times S_{ABC}=\frac29\times S_{ABC}\)

Ta có: \(S_{AGD}+S_{BDE}+S_{GEC}+S_{DEG}=S_{ABC}\)

=>\(S_{DEG}=S_{ABC}-\frac29\times S_{ABC}-\frac29\times S_{ABC}-\frac29\times S_{ABC}=\frac13\times S_{ABC}\)

=>\(S_{DEG}=\frac{450}{3}=150\left(\operatorname{cm}^2\right)\)

Mình học lớp 5 mà chưa học bài này

cô ra thêm bài khó trong giờ học cho mấy bạn giỏi có cái mà làm

Tự làm đi bro