Giải thích các bước giải:
a.Ta có $AH\perp BD\to \widehat{AHB}=\widehat{BAD}=90^o$
Mà $\widehat{ABH}=\widehat{ABD}$
$\to\Delta AHB\sim\Delta DAB(g.g)$
$\to\dfrac{AB}{DB}=\dfrac{HB}{AB}$
$\to AB^2=BH\cdot BD$
Tương tự $AD^2=DH\cdot BD$
$\to \dfrac{HB}{HD}=\dfrac{HB\cdot BD}{HD\cdot BD}=\dfrac{AB^2}{AD^2}=\dfrac{a^2}{b^2}$
b.Ta có $AB\perp AD\to BD^2=AB^2+AD^2=a^2+b^2$
Gọi $HL\cap AB=E\to HE\perp AB$ vì $HL\perp CD, AB//CD$
Mà $HK\perp BC, BE\perp BK$
$\to HEBK$ là hình chữ nhật
$\to BE=HK$
Ta có $\widehat{BEH}=\widehat{BHA}=90^o,\widehat{EBH}=\widehat{ABH}$
$\to\Delta BEH\sim\Delta BHA(g.g)$
$\to \dfrac{BE}{BH}=\dfrac{BH}{BA}$
$\to BE=\dfrac{BH^2}{BA}$
Từ câu a ta có $AB^2=BH\cdot BD$
$\to BH=\dfrac{AB^2}{BD}$
$\to BH^2=\dfrac{AB^4}{BD^2}$
$\to \dfrac{BH^2}{BA}=\dfrac{AB^3}{BD^2}$
$\to BE=\dfrac{AB^3}{BD^2}$
$\to BE=\dfrac{a^3}{a^2+b^2}$
$\to HK=\dfrac{a^3}{a^2+b^2}$
c.Vì $ABCD$ là hình chữ nhật $\to BC=AD=b, DC=AB=a$
Ta có:
$BH=\dfrac{AB^2}{BD}$
$\to BH=\dfrac{a^2}{\sqrt{a^2+b^2}}$
$\to BK^2=BH^2-HK^2$
$\to BK^2=(\dfrac{a^2}{\sqrt{a^2+b^2}})^2-(\dfrac{a^3}{a^2+b^2})^2$
$\to BK^2=\dfrac{a^4b^2}{\left(a^2+b^2\right)^2}$
$\to BK=\dfrac{a^2b}{a^2+b^2}$
$\to CK=BC-BK=b-\dfrac{a^2b}{a^2+b^2}=\dfrac{b^3}{a^2+b^2}$
$\to HC^2=HK^2+CK^2$
$\to HC^2=(\dfrac{a^3}{a^2+b^2})^2+(\dfrac{b^3}{a^2+b^2})^2$
$\to HC^2=\dfrac{a^4-a^2b^2+b^4}{a^2+b^2}$
d.Ta có:
$HK\perp BC\to HK//DC, \widehat{HKC}=\widehat{MDC}=90^o$
Mà $\widehat{KHC}=\widehat{MCD}$
$\to\Delta DMC\sim\Delta KCH(g.g)$
$\to \dfrac{MC}{CH}=\dfrac{DC}{HK}$
$\to MC=\dfrac{DC\cdot CH}{HK}$
$\to MC=\dfrac{a\cdot \sqrt{\dfrac{a^4-a^2b^2+b^4}{a^2+b^2}}}{\dfrac{a^3}{a^2+b^2}}$
$\to MC=\dfrac{\sqrt{a^4-a^2b^2+b^4}\sqrt{a^2+b^2}}{a^2}$
$\to HM=CM-CH=\dfrac{\sqrt{a^4-a^2b^2+b^4}\sqrt{a^2+b^2}}{a^2}-\sqrt{\dfrac{a^4-a^2b^2+b^4}{a^2+b^2}}$
Mà $a=\sqrt{2}, b=1$
$\to HM=\dfrac{\sqrt{(\sqrt{2})^4-(\sqrt{2})^2\cdot 1^2+1^4}\sqrt{(\sqrt{2})^2+1^2}}{(\sqrt{2})^2}-\sqrt{\dfrac{(\sqrt{2})^4-(\sqrt{2})^2\cdot 1^2+1^4}{(\sqrt{2})^2+1^2}}$
$\to HM=\dfrac12$
