a) Xét \(\Delta MNP\) và \(\Delta HMP\) có:
\(\widehat{NMP}=\widehat{MHP}=90^0;\)
\(\widehat{P}\) chung
\(\Rightarrow\Delta MNP\sim\Delta HMP\) (g.g)
b) Xét \(\Delta MNP\) và \(\Delta HNM\) có:
\(\widehat{NHM}=\widehat{NMP}=90^0\); \(\widehat{N}\) chung
\(\Rightarrow\Delta MNP\sim\Delta HNM\) (g.g)
\(\Rightarrow\dfrac{MN}{NP}=\dfrac{NH}{MN}\Rightarrow MN^2=NP.NH\)
c) Tam giác \(MNP\) vuông. Áp dụng định lý Pytago ta có:
\(NP^2=MN^2+MP^2=5^2+6^2=61\) \(\Rightarrow NP=\sqrt{61}\left(cm\right)\)
Do \(\Delta MNP\sim\Delta HNM\Rightarrow\dfrac{MH}{MN}=\dfrac{MP}{NP}\Rightarrow MH=\dfrac{MN.MP}{NP}=\dfrac{30}{\sqrt{61}}\left(cm\right)\)
Lại có \(MN^2=NH.NP\Rightarrow NH=\dfrac{MN^2}{NP}=\dfrac{5^2}{\sqrt{61}}=\dfrac{25}{\sqrt{61}}\left(cm\right)\)
Do \(MK\) là phân giác \(\Rightarrow\dfrac{NK}{MN}=\dfrac{PK}{MP}=\dfrac{NK+PK}{MN+MP}=\dfrac{NP}{MN+MP}=\dfrac{\sqrt{61}}{11}\)
\(\Rightarrow NK=\dfrac{\sqrt{61}}{11}.MN=\dfrac{5\sqrt{61}}{11}\left(cm\right)\)
