Giải thích các bước giải:

a.Xét $\Delta ABD,\Delta ACD$ có:
Chung $AD$

$AB=AC$

$DB=DC$

$\to \Delta ADB=\Delta ADC(c.c.c)$

b.Từ a$\to DC=DB, \widehat{ADB}=\widehat{ADC}$

Mà $\widehat{ADB}+\widehat{ADC}=180^o$

$\to \widehat{ADB}=\widehat{ADC}=90^o$

$\to AD\perp BC=D$ là trung điểm $BC$

$\to AD$ là trung trực $BC$

c.Xét $\Delta ADM,\Delta ADN$ có:

Chung $AD$

$\widehat{DAM}=\widehat{DAN}$

$AM=AN$

$\to \Delta ADM=\Delta ADN(c.g.c)$

$\to \widehat{AND}=\widehat{AMD}=90^o$

$\to DN\perp AC$

d.Xét $\Delta KNE,\Delta KCD$ có:

$KE=KD$

$\widehat{NKE}=\widehat{CKD}$

$KN=KC$

$\to \Delta KNE=\Delta KCD(c.g.c)$

$\to \widehat{KNE}=\widehat{KCD}$

$\to NE//CD$

$\to NE//BC$

Ta có: $AM=AN, AB=AC\to \Delta AMN,\Delta ABC$ cân tại $A$

$\to \widehat{ANM}=90^o-\dfrac12\hat A=\widehat{ACB}\to MN//BC$

VÌ $MN//BC, NE//BC$

$\to M, N, E$ thẳng hàng