a) Theo Thales, có:

$\frac{BM}{MC}=\frac{BK}{CD}$

$\frac{BM}{MC}=\frac{AB}{CP}$

$⇒\frac{AB}{CP}=\frac{BK}{CD}$

$⇔ \frac{BC}{CP}=\frac{BK}{BC}$

Xét $ΔBCK$ và $ΔCPB$, có:

$\frac{BC}{CP}=\frac{BK}{BC}$

$\widehat{KBC}=\widehat{BCP}=90^o$

$⇒ΔBCK\backsimΔCPB$

b) Do $AD//BM$

$⇒\widehat{DAE}=\widehat{AMB}$

Xét $ΔDEA$ và $ΔABM$, có:

$\widehat{DAE}=\widehat{AMB}$

$\widehat{AED}=\widehat{MBA}=90^o$

$⇒ΔDEA\backsimΔABM$

$⇒\frac{DE}{BC}=\frac{AE}{BM}$

$⇔DE.BM=BC.AE$

$⇔ DE.BM=2BM.AE$

$⇒ DE=2AE$

$⇒ AE=DN=NE$

$⇒\frac{AE}{NE}=\frac{AB}{AD}=1$

Do $ΔDEA\backsimΔABM$

$⇒\widehat{ADE}=\widehat{BAM}$

Xét $ΔDNA$ và $ΔAEB$, có:

$\frac{AE}{NE}=\frac{AB}{AD}$

$\widehat{ADN}=\widehat{BAE}$

$⇒ΔDNA\backsimΔAEB$

$⇒\widehat{DNA}=\widehat{AEB}(1)$

Xét $ΔDEB$ có:

$N$ là trung điểm $DE$

$O$ là trung điểm $BD$

$⇒ON$ là đường trung bình $ΔDEB$

$⇒ON//EB$

$⇒\widehat{DNO}=\widehat{DEB}(2)$

Ta có:

$\widehat{DNA}+\widehat{ANO}+\widehat{DNO}=\widehat{AEB}+\widehat{AEN}+\widehat{NEB}=360^o$

Mà $\widehat{DNO}=\widehat{NEB}$, $\widehat{DNA}=\widehat{AEB}$ (CM ở $(1)$ và $(2)$)

$⇒ \widehat{ANO}=\widehat{AEN}=90^o$

$⇒AN⊥NO$

@tuan789