tam giác đồng dạng đây

D

duchuy0405

N

nhuquynhdat

a) Xét $\Delta ABE$ và $\Delta ACF$ có:

$\widehat{AEB}=\widehat{AFC}=90^o$

$\widehat{BAE}=\widehat{CAF}$

$\to \Delta ABE \sim \Delta ACF$
 
N

nhuquynhdat

b) $ \Delta ABE \sim \Delta ACF \to \dfrac{AE}{AF}= \dfrac{BE}{CF}$

$CM: \Delta BDE \sim \Delta CDF (g-g) \to \dfrac{DE}{DF}= \dfrac{BE}{CF}$

$\to \dfrac{AE}{AF}= \dfrac{DE}{DF} \leftrightarrow AE.DF= AF.DE$
 
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