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a/ xét 2 tam giác đói có : \(\frac{AE}{AC}=\frac{AD}{AB}\Leftrightarrow\frac{AE}{AD}=\frac{AC}{AB}\), chung góc A
\(\Rightarrow\Delta AEC\sim\Delta ADB\left(c-g-c\right)\)
b/ xét 2 tam giác đó có: \(\frac{AE}{AC}=\frac{AD}{AB}\), chung góc A
\(\Rightarrow\Delta AED\sim\Delta ACB\left(c-g-c\right)\)
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Xét ΔADE và ΔABC co
AD/AB=AE/AC
góc A chung
=>ΔADE đồng dạng với ΔABC
Xét ΔABE và ΔACD có
AB/AC=AE/AD
góc A chung
=>ΔABE đồng dạng với ΔACD
a: Xét ΔABD và ΔACE có
AB/AC=AD/AE
góc A chung
Do đó: ΔABD\(\sim\)ΔACE
b: ta có: ΔABD\(\sim\)ΔACE
nên \(\dfrac{S_{ABD}}{S_{ACE}}=\left(\dfrac{AB}{AC}\right)^2=\left(\dfrac{5}{7}\right)^2=\dfrac{25}{49}\)
a: Xét ΔABD và ΔACE có
AB/AC=AD/AE
\(\widehat{A}\) chung
Do đó: ΔABD∼ΔACE
b: Xét ΔADE và ΔABC có AD/AB=AE/AC
\(\widehat{A}\) chung
Do đó: ΔADE∼ΔABC