Giải thích các bước giải:

a.Ta có: $\Delta MNP$ vuông tại $M$

$\to NP=\sqrt{MN^2+MP^2}=\sqrt{6^2+(2\sqrt3)^2}=4\sqrt3$

Ta có:

$\cos N=\dfrac{MN}{NP}=\dfrac{2\sqrt3}{4\sqrt3}=\dfrac12$

$\to \hat N=60^o$

$\to \hat P=90^o-\hat N=30^o$

b.Xét $\Delta MDH,\Delta MHN$ có:

Chung $\hat M$

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

$\to \Delta MDH\sim\Delta MHN(g.g)$

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

$\to MH^2=MD.MN$

Tương tự: $MH^2=ME.MP$

$\to MD.MN=ME.MP$

Xét $\Delta MHN,\Delta MHP$ có:

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

$\widehat{HMN}=90^o-\widehat{HMP}=\hat P$

$\to \Delta HMN\sim\Delta HPM(g.g)$

$\to \dfrac{HM}{HP}=\dfrac{HN}{HM}$

$\to HM^2=HN.HP$

$\to HM^2=MN.MP.\dfrac{HN}{NM}.\dfrac{HP}{MP}$

$\to HM^2=MN.MP.\cos\widehat{MNH}.\cos\widehat{MPN}$

$\to HM^2=MN.MP.\sin\widehat{MPN}.\sin\widehat{MNP}$

c.Ta có: $\Delta MNK$ vuông tại $M, MI\perp NK$

$\to NI.NK=NM^2$

Tương tự: $NH.NP=NM^2$

$\to NI.NK=NH.NP$

$\to \dfrac{NI}{NP}=\dfrac{NH}{NK}$

$\to \Delta NHI\sim\Delta NKP(c.g.c)$

$\to\dfrac{HI}{KP}=\dfrac{NH}{NK}$

$\to\dfrac{HI}{KP}=\dfrac{NH}{NM}.\dfrac{NM}{NK}$

$\to\dfrac{HI}{KP}=\sin\widehat{HMN}.\sin\widehat{MKN}$

$\to\dfrac{HI}{KP}=\sin\widehat{MPN}.\sin\widehat{MKN}$