Vì \(\Delta\) ABC vuông tại A => \(\widehat{BAC}=90^o\)
a) Ta có : \(\widehat{BAD}+\widehat{BAC}+\widehat{CAE}=180^o\)
Mà : \(\widehat{BAC}=90^o\) (cmt)
\(\Rightarrow\widehat{BAD}+\widehat{CAE}=90^o\)
Lại có : \(\widehat{CAE}+\widehat{ACE}=90^o\)
\(\Rightarrow\widehat{DAB}=\widehat{ACE}\)
b và c) Xét \(\Delta DAB\) có :
\(\widehat{ADB}+\widehat{DAB}+\widehat{ABD}=180^o\)
Xét \(\Delta ACE\) có :
\(\widehat{ACE}+\widehat{AEC}+\widehat{CAE}=180^o\)
\(\Rightarrow\widehat{ADB}+\widehat{DAB}+\widehat{ABD}=\widehat{ACE}+\widehat{AEC}+\widehat{CAE}\)
Mà : \(\widehat{DAB}=\widehat{ACE}\left(cmt\right);\widehat{BDE}=\widehat{CEA}\left(=90^o\right)\)
\(\Rightarrow\widehat{ABD}=\widehat{CAE}\)
Xét \(\Delta\) ABD và \(\Delta\) ACE có :
$AB = AC (gt)$
\(\widehat{ABD}=\widehat{CAE}\left(cmt\right)\)
\(\widehat{DAB}=\widehat{ACE}\left(cmt\right)\)
\(\Rightarrow\Delta CAE=\Delta ABD\left(gcg\right)\)
$⇒ AD = CE$ ( 2 cạnh tương ứng )
$BD = AE$ ( 2 cạnh tương ứng )
Mà : $DE = AD + AE$
$⇒ DE = BD + CE$
