Giải thích các bước giải:

a.Xét $\Delta ABD,\Delta ACE$ có:

$AB=AE$

$\widehat{BAD}=90^o+\widehat{BAC}=\widehat{EAC}$

$AD=AC$

$\to \Delta ABD=\Delta AEC(c.g.c)$

$\to BD=CE$

b.Xét $\Delta MAB,\Delta MCN$ có:

$MA=MN$

$\widehat{AMB}=\widehat{CMN}$

$MB=MC$

$\to \Delta MAB=\Delta MNC(c.g.c)$

$\to AB=CN,\widehat{MAB}=\widehat{MNC}\to AB//CN$

$\to \widehat{ACN}=180^o-\widehat{BAC}=360^o-\widehat{BAE}-\widehat{CAD}-\widehat{BAC}=\widehat{DAE}$

Mà $CN=AB=AE, AC=AD$

$\to \Delta ADE=\Delta CAN(c.g.c)$

$\to DE=AN$