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nhuquynhdat

CM: $AI=CK $ ( tự CM)

Xét $\Delta ABK$ có $ME//BK$

$\to \dfrac{AB}{AE}= \dfrac{AK}{AM}$

Tương tự có $\dfrac{AD}{AF}= \dfrac{AI}{AM}$

$\dfrac{AB}{AE}+\dfrac{AD}{AF}= \dfrac{AK}{AM}+ \dfrac{AI}{AM}= \dfrac{AK+AI}{AM}= \dfrac{AI+IK+CK}{AM}= \dfrac{AC}{AM}$
 
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