Giải thích các bước giải:

Bài 2:

a.Ta có $AB=AC, BD=CE\to AD=AB+BD=AC+CE=AE\to\Delta ABC, ADE$ cân

$\to\widehat{ABC}=90^o-\dfrac{\widehat{BAC}}{2}=\widehat{ADE}\to BC//DE$

b.Ta có : $\widehat{DMB}=\widehat{NCE}=90^o, \widehat{DBM}=\widehat{ABC}=\widehat{ACB}=\widehat{ECN}$

Kết hợp $BD=CE\to\Delta BMD=\Delta CNE(g.c.g)\to DM=EN, MB=CN$

c.Lại có $\widehat{ABM}=180^o-\widehat{ABC}=180^o-\widehat{ACB}=\widehat{ACN}, AB=AC$

$\to\Delta ABM=\Delta ACN(c.g.c)\to AM=AN\to\Delta AMN$ cân tại A

d.Gọi $IB\perp AM=F,IC\perp AN=G, AI\perp CB=H$

$\to \widehat{BFM}=\widehat{CGN}=90^o$ mà $\widehat{FMB}=\widehat{CNG}, MB=CN$

$\to\Delta FMB=\Delta GNC(g.c.g)$

$\to MF=NG, \widehat{FBM}=\widehat{GCN}, BF=CG$

$\to AF=AG, \widehat{IBC}=\widehat{ICB}\to IB=IC$

$\to IF=IG\to \Delta IAF=\Delta IAG, \Delta IAB=\Delta IAC(c.c.c)$

$\to \widehat{BAI}=\widehat{IAC}, \widehat{FAI}=\widehat{IAG}\to AI$ là phân giác $\widehat{BAC}, \widehat{MAN}$