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23 tháng 2 2020

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

\(\Rightarrow\Delta AED\sim\Delta ACB\left(c-g-c\right)\)

24 tháng 4 2021

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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

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 \(\dfrac{S_{ABD}}{S_{ACE}}=\left(\dfrac{AB}{AC}\right)^2=\left(\dfrac{5}{7}\right)^2=\dfrac{25}{49}\)

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