toan8

N

nhuquynhdat

Kẻ $DE//KM; BF//KM$

CM: $\Delta ADE= \Delta CBF (g-c-g) \to AE= CF$

Xét $\Delta ADG$ có $DE//KM \to DE//KG$

$\to \dfrac{AD}{AK}= \dfrac{AE}{AG}$

Xét $\Delta CBF$ có $BF//KM \to BF//GM$

$\to \dfrac{AB}{AM}= \dfrac{AF}{AG}$

$\to \dfrac{AD}{AK}+ \dfrac{AB}{AM}=\dfrac{AE}{AG}+\dfrac{AF}{AG}= \dfrac{AE+AF}{AG}= \dfrac{AE+CF+EF}{AG}= \dfrac{AC}{AG}$
 
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