a. Ta có :
\(AB=BD\left(gt\right)\)
\(\Leftrightarrow\Delta ABD\) cân tại B
\(\Leftrightarrow\widehat{BAD}=\widehat{D1}\)
Lại có : \(\widehat{BAD}+\widehat{A3}=90^0\)
\(\Leftrightarrow\widehat{D1}+\widehat{A3}=90^0\)
Mà \(\widehat{A2}+\widehat{D1}=90^0\)
\(\Leftrightarrow\widehat{A2}=\widehat{A3}\)
Xét \(\Delta HAD,\Delta EAD\) có :
\(\left\{{}\begin{matrix}AH=AE\left(gt\right)\\\widehat{A2}=\widehat{A3}\\ADchung\end{matrix}\right.\)
\(\Leftrightarrow\Delta HAD=\Delta EAD\left(c-g-c\right)\)
\(\Leftrightarrow\widehat{AHD}=\widehat{AED}=90^0\)
\(\Leftrightarrow AE\perp EC\left(đpcm\right)\)
b/ Xét \(\Delta DEC\) vuông tại E :
\(\Rightarrow DC>EC\)
Ta có : \(BC+AH=BD+DC+AH=AB+DC+AH>AB+EC+AE=AB+AC\left(đpcm\right)\)
