Giải thích các bước giải:
Gọi F đối xứng với A qua M $\to AF=2AM$
Mà MA=MF,MB=MC $,\widehat{AMB}=\widehat{CMF}\to\Delta AMB=\Delta FMC(c.g.c)$
$\to CF=AB=AD, \widehat{BAM}=\widehat{MFC}\to AB//CF$
$\to \widehat{FCA}+\widehat{BAC}=180^o$
Mà $\widehat{DAE}+\widehat{ADB}+\widehat{BAC}+\widehat{CAE}=360^o$
$AD\perp AB, AE\perp AC\to \widehat{DAE}+\widehat{BAC}=180^o$
$\to \widehat{DAE}=\widehat{ACF}$
Lại có : $AE=AC\to \Delta ADE=\Delta CFA(c.g.c)\to DE=AF=2AM\to AM=\dfrac{DE}2$
Gọi $AM\cap DE= G$
$\to\widehat{DAG}+\widehat{BAM}=90^o$
Mà $\widehat{BAM}=\widehat{MFC}=\widehat{AFC}=\widehat{ADE}$
$\to\widehat{DAG}+\widehat{ADG}=90^o\to AM\perp DE$
