Xét \(\Delta AEC\&\Delta ADB\\ \) có:
\(\widehat{A}=\widehat{A}\\ \widehat{E}=\widehat{D}=90^o\\ \Rightarrow\Delta AEC\sim\Delta ADB\left(đpcm\right)\)
b) vì\(\Delta AEC\sim\Delta ADB\Leftrightarrow\dfrac{AB}{AE}=\dfrac{AC}{AD}\Leftrightarrow\dfrac{3}{AE}=\dfrac{5}{2}\Rightarrow AE=\dfrac{3\cdot2}{5}=1.2cm\)
a) Xét ∆ADB và ∆ACE có:
∠ADB = ∠ACE = 90⁰
∠A chung
⇒ ∆ADB ∽ ∆ACE (g-g)
b) Do ∆ADB ∽ ∆ACE (cmt)
⇒ AD/AC = AB/AE
⇒ AE = AB.AC/AD
= 2.3/5
= 1,2 (cm)