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T

thienbinhgirl

Ta có: ΔAEB∼ΔMED(g.g)→MEEA=DMAB\Delta AEB\sim \Delta MED (g.g)\rightarrow \frac{ME}{EA}=\frac{DM}{AB}
ΔCFM∼ΔAFB(g.g)→MFFB=MCAB\Delta CFM\sim \Delta AFB (g.g)\rightarrow \frac{MF}{FB}=\frac{MC}{AB}
Mà DM=MC (gt) \Rightarrow MEEA=MFFB\frac{ME}{EA}=\frac{MF}{FB} \Rightarrow EF//AB
 
T

thienbinhgirl

CD=12cm→DM=MC=6cm→DMAB=MCAB=67,5=45→EFAB=45→EF=7,5.45=6cmCD=12cm \rightarrow DM=MC=6cm \rightarrow \frac{DM}{AB}=\frac{MC}{AB}=\frac{6}{7,5}=\frac{4}{5}\rightarrow \frac{EF}{AB}=\frac{4}{5}\rightarrow EF=\frac{7,5.4}{5}=6 cm
 
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