Giải thích các bước giải:

a.Xét $\Delta ABE,\Delta ACF$ có:
Chung $\hat A$

$\widehat{AEB}=\widehat{AFC}(=90^o)$

$\to \Delta ABE\sim\Delta ACF(g.g)$

$\to\dfrac{AB}{AC}=\dfrac{AE}{AF}$

$\to AE\cdot AC=AF\cdot AB$

$\to \dfrac{AE}{AB}=\dfrac{AF}{AC}$

Mà $\widehat{EAF}=\widehat{BAC}$

$\to \Delta ABC\sim\Delta AEF(c.g.c)$

b.Ta có: $H, K$ đối xứng qua $M\to M$ là trung điểm $HK$

               $M$ là trung điểm $BC$

$\to BHCK$ là hình bình hành

$\to CK//BH, CH//BK\to CK\perp AC, BK\perp AB$

Mặt khác $BK=CH, BH=CK$

Xét $\Delta BDH,\Delta DAC$ có:

$\widehat{HDB}=\widehat{ADC}(=90^o)$

$\widehat{HBD}=90^o-\widehat{BHD}=90^o-\widehat{AHE}=\widehat{HAE}=\widehat{DAC}$

$\to \Delta DBH\sim\Delta DAC(g.g)$

$\to \dfrac{DB}{DA}=\dfrac{BH}{AC}$

$\to \dfrac{BD}{DA}=\dfrac{CK}{AC}$

Mà $\widehat{ADB}=\widehat{ACK}(=90^o)$

$\to \Delta ABD\sim\Delta AKC(c.g.c)$

$\to \widehat{KAC}=\widehat{BAD}$

Gọi $AK\cap EF=G\to \widehat{GAE}=\widehat{BAD}$

Từ câu a $\to\widehat{AEF}=\widehat{ABC}\to \widehat{AEG}=\widehat{ABD}=\widehat{AKC}$

$\to \Delta AGE\sim\Delta ACK(g.g)$

$\to \widehat{AGE}=\widehat{ACK}=90^o$

$\to AK\perp EF$

c.Xét $\Delta NBF,\Delta NCE$ có:

Chung $\hat N$

$\widehat{NFB}=\widehat{AFE}=\widehat{ACB}=\widehat{NCE}$

$\to\Delta NBF\sim\Delta NEC(g.g)$

$\to \dfrac{NB}{NE}=\dfrac{NF}{NC}$

$\to NB\cdot NC=NE\cdot NF$

Tương tự câu a chứng minh được $\Delta BDF\sim\Delta BAC(g.g)\to \widehat{FDB}=\widehat{BAC}$

Ta có: $\Delta EBC$ vuông tại $E, M$ là trung điểm $BC$

$\to ME=MB=MC=\dfrac12BC$

$\to\Delta MCE$ cân tại $M$

$\to \widehat{MEC}=\widehat{MCE}=\widehat{ACB}$

$\to \widehat{NEM}=180^o-\widehat{AEF}-\widehat{MEC}=180^o-\widehat{ABC}-\widehat{ACB}=\widehat{BAC}=\widehat{FDB}=\widehat{FDN}$

Do $\widehat{FND}=\widehat{ENM}$

$\to \Delta NDF\sim\Delta NEM(g.g)$

$\to \dfrac{ND}{NE}=\dfrac{NF}{NM}$

$\to NE\cdot NF=ND\cdot NM$

$\to ND\cdot NM=NB\cdot NC$

Ta có: $NB+NC=(NM-MB)+(NM+MC)=2MN+(MC-MB)=2MN$ vì $MB=MC$

$\to \dfrac{NB+NC}{NB\cdot NC}=\dfrac{2MN}{ND\cdot NM}$

$\to \dfrac1{NC}+\dfrac1{NB}=\dfrac2{ND}$