[Toán 8] Hình học

S

shinbc

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P

pinkylun

$\triangle{ABK}$ có $DM//AB$

$=>\dfrac{DM}{AB}=\dfrac{MB}{MK}$

Tương tự $\triangle{ABI}$ có $MC//AB$

$=>\dfrac{CM}{AB}=\dfrac{AM}{MI}$

Mà $DM=CM$

$=>\dfrac{MB}{MK}=\dfrac{AM}{MI}$

Theo định lí Ta lét đảo =>đpcm
 
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