Đáp án:

 

Giải thích các bước giải:

 

Giải thích các bước giải:

a.Ta có $ABCD$ là hình vuông $\to AC\perp BD=O$

Mà $\widehat{IOM}=90^o$

$\to\widehat{cOB}=\widehat{IOM}(=90^o)$ 

$\to\widehat{COB}-\widehat{BOM}=\widehat{IOM}-\widehat{BOM}$ 

$\to\widehat{BOI}=\widehat{MOC}$

Mà $OB=OC,\widehat{IBO}=\widehat{OBA}=45^o=\widehat{OCB}=\widehat{OCM}$

$\to\Delta BIO=\Delta CMO(g.c.g)$

$\to S_{BOI}=S_{COM}$

$\to S_{BIOM}=S_{BOI}+S_{OBM}=S_{COM}+S_{OBM}=S_{BOC}=\dfrac14S_{ABCD}=\dfrac14a^2$

b.Ta có:
$CM//AD\to\dfrac{CM}{AD}=\dfrac{NC}{ND}$

$\to\dfrac{CM}{BC}=\dfrac{NC}{ND}$

$\to\dfrac{CM}{BC-CM}=\dfrac{NC}{ND-NC}$

$\to\dfrac{CM}{BM}=\dfrac{NC}{CD}$

$\to\dfrac{CM}{BM}=\dfrac{NC}{BC}$

Xét $\Delta BKC,\Delta CKN$ có:
$\widehat{BKC}=\widehat{CKN}=90^o,\widehat{BCK}=90^o-\widehat{KCN}=\widehat{KNC}$

$\to\Delta BCK\sim\Delta CNK(g.g)$

$\to\dfrac{BC}{CN}=\dfrac{BK}{CK}$

$\to\dfrac{NC}{BC}=\dfrac{CK}{BK}$

$\to \dfrac{CM}{BM}=\dfrac{KC}{KB}$

$\to KM$ là phân giác $\widehat{BKC}$

$\to\widehat{BKM}=\dfrac12\widehat{BKC}=45^o=\widehat{BCO}$

c.Kẻ $AE\perp AN, E\in CD$

$\to\widehat{EAD}=90^o-\widehat{DAM}=\widehat{MAB}$

Mà $AD=AB,\widehat{ABM}=\widehat{ADE}=90^o$

$\to \Delta ABM=\Delta ADE(c.g.c)$ 

$\to AM=AE$

Ta có: $AD\perp NE$

$\to\widehat{ADE}=\widehat{ADN}=90^o,\widehat{EAD}=90^o-\widehat{DAN}=\widehat{AND}$

$\to\Delta ADE\sim\Delta NDA(g.g)$

$\to\dfrac{AD}{DN}=\dfrac{AE}{NA}=\dfrac{DE}{DA}$

$\to AD^2=DE.DN$

Lại có $\widehat{AED}=\widehat{AEN},\widehat{ADE}=\widehat{EAN}=90^o$

$\to\Delta ADE\sim\Delta NAE(g.g)$

$\to\dfrac{AE}{EN}=\dfrac{DE}{AE}$

$\to AE^2=ED.EN$

Tương tự $NA^2=ND.NE$

$\to \dfrac{1}{AE^2}+\dfrac{1}{AN^2}=\dfrac{1}{ED.EN}+\dfrac{1}{ND.NE}$

$\to \dfrac{1}{AE^2}+\dfrac{1}{AN^2}=\dfrac1{EN}.(\dfrac{1}{ED}+\dfrac{1}{ND})$

$\to \dfrac{1}{AE^2}+\dfrac{1}{AN^2}=\dfrac1{EN}.\dfrac{ND+ED}{ED.ND}$

$\to \dfrac{1}{AE^2}+\dfrac{1}{AN^2}=\dfrac1{EN}.\dfrac{EN}{AD^2}$

$\to \dfrac{1}{AE^2}+\dfrac{1}{AN^2}=\dfrac{1}{AD^2}$

$\to \dfrac{1}{AM^2}+\dfrac{1}{AN^2}=\dfrac{1}{CD^2}$