Toán hình khó, cần gấp

N

nhuquynhdat

a) Xét $\Delta EDM$ có : $\dfrac{EM}{EA}=\dfrac{DM}{AB}$

Xét $\Delta FCM$ có : $\dfrac{FM}{FB}=\dfrac{MC}{AB}$

Do $DM = CM \to \dfrac{DM}{AB}= \dfrac{MC}{AB}$

$\to \dfrac{EM}{EA}= \dfrac{FM}{FB} \to EF//AB$( Ta- let đảo)
 
T

trinhminh18

Theo định lý Ta-lét
$\dfrac{EF}{AB}$=$\dfrac{EM}{AM}$
mà$\dfrac{EM}{AM}$=$\dfrac{DM}{AB+DM}$
\Rightarrow$\dfrac{EF}{7,5}$=$\dfrac{6}{6+7,5}$
\RightarrowEF=$\dfrac{10}{3}$
 
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