c. -Xét △ADC có: OM//DC (gt).

\(\Rightarrow\dfrac{MO}{DC}=\dfrac{AO}{AC}\) (định lí Ta-let).

\(\Rightarrow\dfrac{DC}{MO}=\dfrac{AC}{AO}\)

\(\Rightarrow\dfrac{DC}{OM}-1=\dfrac{OC}{AO}\) (1).

-Xét △BDC có: ON//DC (gt).

\(\Rightarrow\dfrac{ON}{DC}=\dfrac{BO}{BD}\) (định lí Ta-let).

\(\Rightarrow\dfrac{DC}{ON}=\dfrac{BD}{BO}\)

\(\Rightarrow\dfrac{DC}{ON}-1=\dfrac{OD}{BO}\)

-Xét △ABO có: AB//DC (gt).

\(\Rightarrow\dfrac{OD}{BO}=\dfrac{OC}{OA}=\dfrac{DC}{AB}\) (3)

-Từ (1), (2),(3) suy ra:

\(\dfrac{DC}{OM}-1=\dfrac{DC}{ON}-1=\dfrac{DC}{AB}\)

\(\Rightarrow\dfrac{DC}{OM}=\dfrac{DC}{ON}=\dfrac{DC}{AB}+1=\dfrac{AB+DC}{AB}\)

\(\Rightarrow\dfrac{1}{OM}=\dfrac{1}{ON}=\dfrac{AB+DC}{AB.DC}=\dfrac{1}{AB}+\dfrac{1}{CD}\)

a: Xét ΔAOB và ΔCOD có 

\(\widehat{OAB}=\widehat{OCD}\)

\(\widehat{AOB}=\widehat{COD}\)

Do đó: ΔAOB∼ΔCOD

Suy ra: \(\dfrac{OA}{OC}=\dfrac{OB}{OD}=\dfrac{AB}{CD}\)

hay \(OA\cdot OD=OB\cdot OC\)

b: \(\dfrac{OA}{OC}=\dfrac{AB}{CD}\)

\(\Leftrightarrow OA=\dfrac{1}{2}\cdot6=3\left(cm\right)\)

a: Xét ΔOAB và ΔOCD có

góc OAB=góc OCD

góc AOB=góc COD
=>ΔOAB đồng dạng vơi ΔOCD

=>OA/OC=OB/OD=AB/CD

=>OA*OD=OB*OC

b: OA/OC=AB/CD

=>OA/6=5/10=1/2

=>OA=3cm

Xet ΔADC có OE//DC

nên OE/DC=AO/AC

=>OE/10=3/(3+6)=3/9=1/3

=>OE=10/3cm

a)  Xét  \(\Delta OAB\)và   \(\Delta OCD\)có:

      \(\widehat{OAB}=\widehat{OCD}\) (slt)

      \(\widehat{OBA}=\widehat{ODC}\) (slt)

suy ra:   \(\Delta OAB~\Delta OCD\)  (g.g)

\(\Rightarrow\)\(\frac{OA}{OC}=\frac{OB}{OD}\)

\(\Rightarrow\)\(OA.OD=OB.OC\)

b)  \(\Delta OAB~\Delta OCD\)  

\(\Rightarrow\)\(\frac{OA}{AC}=\frac{AB}{CD}\)

\(\Rightarrow\)\(OA=\frac{OC.AB}{CD}=3\)

\(\Rightarrow\)\(AC=OA+OC=9\)

\(\Delta AEO~\Delta ADC\)  ( do OE // DC )

\(\Rightarrow\)\(\frac{OE}{DC}=\frac{OA}{AC}\)  \(\Rightarrow\) \(OE=\frac{OA.DC}{AC}=\frac{10}{3}\)

Chọn C

a: Xét ΔOAB và ΔOCD có

góc OAB=góc OCD

góc AOB=góc COD

=>ΔOAB đồng dạng với ΔOCD
=>AB/CD=OA/OC=OB/OD

=>5/CD=1/2

=>CD=10cm và OA*OD=OB*OC

b: Xét ΔOHA vuông tại H và ΔOKC vuông tại K có

góc AOH=góc KOC

=>ΔOHA đồng dạng với ΔOKC

=>OH/OK=OA/OC=1/2

c: AE/AD+CF/BC

=AE/AD+1-BF/BC

=1

Xét ΔOAB và ΔOCD có

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

\(\widehat{AOB}=\widehat{COD}\)

Do đó: ΔOAB đồng dạng với ΔOCD

=>\(\dfrac{OA}{OC}=\dfrac{OB}{OD}=\dfrac{AB}{CD}=\dfrac{6}{9}=\dfrac{2}{3}\)

\(\dfrac{OA}{OC}=\dfrac{2}{3}\)

=>\(OC=1,5OA\)

\(\dfrac{OB}{OD}=\dfrac{2}{3}\)

=>\(OD=3\cdot\dfrac{OB}{2}=1,5OB\)

AO+OC=AC

=>1,5OA+OA=OC

=>OC=2,5OA

=>\(\dfrac{OC}{OA}=2,5=\dfrac{5}{2}\)

=>\(\dfrac{OA}{OC}=\dfrac{2}{5}\)

OB+OD=BD

=>BD=1,5OB+OB=2,5OB

=>\(\dfrac{OB}{BD}=\dfrac{2}{5}\)

Xét ΔADC có MO//DC

nên \(\dfrac{MO}{DC}=\dfrac{AO}{AC}\)

=>\(\dfrac{MO}{9}=\dfrac{2}{5}=0,4\)

=>MO=0,4*9=3,6(cm)

Xét ΔBDC có ON//DC

nên \(\dfrac{ON}{DC}=\dfrac{BO}{BD}\)

=>\(\dfrac{ON}{9}=\dfrac{2}{5}\)

=>ON=0,4*9=3,6(cm)

MN=MO+ON

=3,6+3,6

=7,2(cm)