Giải thích các bước giải:

a.Xét $\Delta MNH,\Delta MHD$ có:

Chung $\hat M$

$\widehat{MHN}=\widehat{MDH}(=90^o)$

$\to \Delta MNH\sim\Delta MHD(g.g)$

$\to \dfrac{MN}{MH}=\dfrac{MH}{MD}$

$\to MH^2=MD\cdot MN$

b.Tương tự câu a chứng minh được: $MH^2=ME\cdot MP$

$\to MD\cdot MN=ME\cdot MP$

c.Xét $\Delta MDE,\Delta MNP$ có:

Chung $\hat M$

$\dfrac{ME}{MN}=\dfrac{MD}{MP}$ vì $ MD\cdot MN=ME\cdot MP$

$\to \Delta MED\sim\Delta MNP(c.g.c)$

$\to \widehat{MDE}=\widehat{MPN}$

d.Xét $\Delta MDP,\Delta MNE$ có:

Chung $\hat M$

$\dfrac{MD}{MP}=\dfrac{ME}{MN}\to \dfrac{MD}{ME}=\dfrac{MP}{MN}$

$\to \Delta MDP\sim\Delta MEN(c.g.c)$

$\to \widehat{MPD}=\widehat{MNE}$

Trên $DP$ lấy điểm $A$ sao cho $\widehat{AEP}=\widehat{DEN}$

Mà $\widehat{END}=\widehat{MNE}=\widehat{MPD}=\widehat{EPA}$

$\to \Delta EDN\sim\Delta EAP(g.g)$

$\to \dfrac{EN}{EP}=\dfrac{DN}{AP}$

$\to EN\cdot AP=EP\cdot DN$

Mặt khác $\widehat{EAP}=\widehat{EDN}$

$\to \widehat{EAD}=180^o-\widehat{EAP}=180^o-\widehat{EDN}=\widehat{MDE}=\widehat{MPN}=\widehat{EPN}$

Gọi $NE\cap PD=B$

$\to \widehat{DBN}=\widehat{EBP}$

Mà $\widehat{BND}=\widehat{ENM}=\widehat{MPD}=\widehat{EPB}$

$\to \Delta BDN\sim\Delta BEP(g.g)$

$\to \dfrac{BD}{BE}=\dfrac{BN}{BP}\to \dfrac{BD}{BN}=\dfrac{BE}{BP}$

Mà $\widehat{DBE}=\widehat{NBP}$

$\to \Delta BDE\sim\Delta BNP(c.g.c)$

$\to \widehat{BDE}=\widehat{BNP}$

$\to \widehat{EDA}=\widehat{ENP}$

Kết hợp $\widehat{EAD}=\widehat{EPN}$

$\to \Delta EDA\sim\Delta ENP(g.g)$

$\to \dfrac{ED}{EN}=\dfrac{AD}{NP}$

$\to ED\cdot NP=EN\cdot DA$

$\to DE\cdot NP+ND\cdot EP=EN\cdot DA+PA\cdot EN=EN\cdot (DA+AP)=EN\cdot DP$