Giải thích các bước giải:

i.Ta có: $\Delta MNF$ vuông tại $M, MO\perp NF$

$\to NF=\sqrt{MF^2+MN^2}=15$

$\to \sin\widehat{MFN}=\dfrac{MN}{NF}=\dfrac35\to \widehat{MFN}\approx 36.87^o$

$\to \widehat{MNF}=90^o-\widehat{MFN}=53.13^o$

ii.Vì $\Delta MNF$ vuông tại $M, MO\perp NF$

$\to MO.NF=MN.MF$

$\to MO=\dfrac{MN.MF}{FN}=\dfrac{9\cdot 12}{15}=7.2$

$\to FO=\sqrt{MF^2-MO^2}=\sqrt{12^2-7.2^2}=9.6$

iii.Ta có: $NM\perp MF, MF\perp FE, NH\perp FE$

$\to NHFM$ là hình chữ nhật

$\to NH=MF=12, FH=MN=9$

Vì $\Delta MFE$ vuông tại $M, FO\perp ME$

$\to MO.ME=MF^2$

$\to ME=\dfrac{MF^2}{MO}=\dfrac{12^2}{7.2}=20$

$\to EF=\sqrt{ME^2-MF^2}=16$

 $\to S_{NEF}=\dfrac12NH\cdot EF=\dfrac12\cdot 12\cdot 16=96$

Ta có: $S_{NHF}=S_{MNF}=\dfrac12\cdot MN\cdot MF=\dfrac12\cdot 9\cdot 12=54$

           $\dfrac{FO}{ON}=\dfrac{FE}{MN}=\dfrac{16}{9}$

$\to \dfrac{FO}{FO+ON}=\dfrac{16}{16+9}$

$\to \dfrac{FO}{FN}=\dfrac{16}{25}$

$\to S_{OHF}=\dfrac{!6}{25}S_{HNF}=\dfrac{16}{25}\cdot 54=\frac{864}{25}$

b.Xét $\Delta MNF,\Delta MFE$ có:

$\widehat{NMF}=\widehat{MFE}(=90^o)$

$\widehat{MNF}=90^o-\widehat{MFN}=90^o-\widehat{MFO}=\widehat{FMO}=\widehat{FME}$

$\to \Delta MNF\sim\Delta FME(g.g)$

$\to \dfrac{MF}{FE}=\dfrac{MN}{MF}$

$\to MF^2=MN.EF$