Đáp án:

a) $S_{ABCD}=64cm^2, S_{AMCD}=40cm^2$

b) $AN=\dfrac{8}{3}cm$

c) $\triangle AMN\backsim\triangle BMC$

d) $CD^2=ND.MB$

e) $HK=3,6cm$

Giải thích các bước giải:

a)

$S_{ABCD}=8^2=64(cm^2)\\S_{AMCD}=\dfrac{(2+8).8}{2}=40(cm^2)$

b)

Xét $\triangle CDN$ và $\triangle MBC$:

$\widehat{CDN}=\widehat{MBC}\,\,\,(=90^o)$

$\widehat{CND}=\widehat{MCB}$ (so le trong)

$\to\triangle CDN\backsim\triangle MBC$ (g.g)

$\to\dfrac{CD}{ND}=\dfrac{MB}{CB}\\\to ND=\dfrac{CD.CB}{MB}=\dfrac{8.8}{6}=\dfrac{32}{3}(cm)\\\to AN=ND-AD=\dfrac{32}{3}-8=\dfrac{8}{3}(cm)$

c)

Xét $\triangle AMN$ và $\triangle BMC$:

$\widehat{NAM}=\widehat{CBM}\,\,\,(=90^o)$

$\widehat{AMN}=\widehat{BMC}$ (đối đỉnh)

$\to\triangle AMN\backsim\triangle BMC$ (g.g)

d)

$\triangle CDN\backsim\triangle MBC$ (cmt)

$\to\dfrac{CD}{ND}=\dfrac{MB}{CB}$

Mà $CB=CD$ (tứ giác ABCD là hình vuông)

$\to\dfrac{CD}{ND}=\dfrac{MB}{CD}\\\to CD^2=ND.MB$

e)

$\triangle CDN$ vuông tại D:

$DC^2+DN^2C=CN^2$ (định lý Pytago)

$\to CN=\sqrt{DC^2+DN^2}=\sqrt{8^2+\left(\dfrac{32}{3}\right)^2}=\dfrac{40}{3}(cm)$

$S_{\triangle CHN}=\dfrac{1}{2}.DC.DN=\dfrac{1}{2}.DH.CN\\\to DC.DN=DH.CN\\\to DH=\dfrac{DC.DN}{CN}=\dfrac{8.\dfrac{32}{3}}{\dfrac{40}{3}}=6,4(cm)$

$\triangle DHN$ vuông tại H:

$HN^2+HD^2=ND^2$ (định lý Pytago)

$\to HN=\sqrt{ND^2-HD^2}=\sqrt{\left(\dfrac{32}{3}\right)^2-6,4^2}=\dfrac{128}{15}(cm)$

$\to CH=CN-HN=\dfrac{40}{3}-\dfrac{128}{15}=4,8(cm)$

Xét $\triangle DHN$ và $\triangle KHC$:

$\widehat{DHN}=\widehat{KHC}$ (đối đỉnh)

$\widehat{DNH}=\widehat{KCH}$ (so le trong)

$\to\triangle DHN\backsim\triangle KHC$ (g.g)

$\to\dfrac{DH}{HN}=\dfrac{KH}{HC}\\\to KH=\dfrac{DH.HC}{HN}=\dfrac{6,4.4,8}{\dfrac{128}{15}}=3,6(cm)$