Giải thích các bước giải:
a.Xét $\Delta MNP,\Delta HMN$ có:
Chung $\hat N$
$\widehat{NMP}=\widehat{MHN}(=90^o)$
$\to\Delta MNP\sim\Delta HNM(g.g)$
b.Ta có $\Delta MNP$ vuông tại $M, MH\perp NP$
$\to MH\cdot NP=MN\cdot MP(=2S_{MNP})$
c.Vì $\Delta MNP$ vuông tại $M$
$\to NP^2=MN^2+MP^2=625$
$\to NP=25$
Do $MD$ là phân giác $\hat M$
$\to \dfrac{DN}{DP}=\dfrac{MN}{MP}=\dfrac34$
$\to \dfrac{DN}{DN+DP}=\dfrac3{3+4}$
$\to\dfrac{DN}{NP}=\dfrac37$
$\to DN=\dfrac37NP$
$\to DN=\dfrac{75}7$
Từ câu b $\to MH=\dfrac{MN\cdot MP}{NP}=12$
$\to NH=\sqrt{MN^2-MH^2}=9$
$\to HD=DN-NH=\dfrac{12}7$
d.Gọi $MK\cap NP=A$
Xét $\Delta AKN,\Delta AMH$ có:
Chung $\hat A$
$\widehat{AKN}=\widehat{AHM}(=90^o)$
$\to\Delta AKN\sim\Delta AHM(g.g)$
$\to \dfrac{AK}{AH}=\dfrac{AN}{AM}$
$\to\dfrac{AN}{AK}=\dfrac{AM}{AH}$
Mà $\widehat{MAN}=\widehat{KAH}$
$\to\Delta ANM\sim\Delta AKH(c.g.c)$
$\to\widehat{AMN}=\widehat{AHK}$
$\to \widehat{NHK}=\widehat{KMN}=90^o-\widehat{KMI}=\widehat{KIM}=\widehat{NIP}$
Lại có $\widehat{KNH}=\widehat{INP}$
$\to\Delta NHK\sim\Delta NIP(g.g)$
