a: Ta có: M là trung điểm của CD
=>\(MC=MD=\dfrac{CD}{2}=10\left(cm\right)\)
Xét ΔEAB và ΔEMD có
\(\widehat{EAB}=\widehat{EMD}\)(hai góc so le trong, AB//MD)
\(\widehat{AEB}=\widehat{MED}\)(hai góc đối đỉnh)
Do đó; ΔEAB~ΔEMD
=>\(\dfrac{EA}{EM}=\dfrac{AB}{DM}=\dfrac{15}{10}=\dfrac{3}{2}\left(1\right)\)
Xét ΔFAB và ΔFCM có
\(\widehat{FAB}=\widehat{FCM}\)(hai góc so le trong, AB//CM)
\(\widehat{AFB}=\widehat{CFM}\)(hai góc đối đỉnh)
Do đó; ΔFAB~ΔFCM
=>\(\dfrac{FB}{FM}=\dfrac{AB}{CM}=\dfrac{15}{10}=\dfrac{3}{2}\left(2\right)\)
Từ (1),(2) suy ra \(\dfrac{AE}{EM}=\dfrac{BF}{FM}=\dfrac{3}{2}\)
=>\(\dfrac{EM}{AE}=\dfrac{MF}{BF}=\dfrac{2}{3}\)
Xét ΔMAB có \(\dfrac{ME}{EA}=\dfrac{MF}{FB}\)
nên EF//AB
b: ta có: \(\dfrac{ME}{EA}=\dfrac{2}{3}\)
=>\(\dfrac{EA}{EM}=\dfrac{3}{2}\)
=>\(\dfrac{EA}{EM}+1=\dfrac{5}{2}\)
=>\(\dfrac{EA+EM}{EM}=\dfrac{5}{2}\)
=>\(\dfrac{AM}{ME}=\dfrac{5}{2}\)
=>\(\dfrac{ME}{MA}=\dfrac{2}{5}\)
Xét ΔMAB có EF//AB
nên \(\dfrac{ME}{MA}=\dfrac{EF}{AB}\)
=>\(\dfrac{EF}{15}=\dfrac{2}{5}\)
=>\(EF=15\cdot\dfrac{2}{5}=6\left(cm\right)\)
