Giải thích các bước giải:

a.Xét $\Delta ABD,\Delta ACE$ có:

Chung $\hat A$

$\widehat{ADB}=\widehat{AEC}(=90^o)$

$\to\Delta ABD\sim\Delta  ACE(g.g)$

b.Xét $\Delta CHD,\Delta CAE$ có:

Chung $\hat C$

$\widehat{CDH}=\widehat{CEA}(=90^o)$

$\to\Delta CDH\sim\Delta CEA(g.g)$

$\to\dfrac{CD}{CE}=\dfrac{CH}{CA}$

$\to CH.CE=CD.CA$

c.Ta có $DI\perp CE, CE\perp AB$

$\to \dfrac{CD}{CA}=\dfrac{CI}{CE}\to CA.CI=CD.CE$

  Ta có $EK\perp AC\to EK//BD$

$\to \dfrac{CD}{CK}=\dfrac{CH}{CE}\to CD.CE=CH.CK$

$\to CA.CI=CH.CK$

$\to \dfrac{CI}{CH}=\dfrac{CK}{CA}$

$\to KI//AH$

d.Xét $\Delta CID,\Delta CKE$ có:

Chung $\hat C$

$\widehat{CID}=\widehat{CKE}(=90^o)$

$\to\Delta CDI\sim\Delta CEK(g.g)$

$\to\dfrac{CD}{CE}=\dfrac{CI}{CK}\to \dfrac{CD}{CI}=\dfrac{CE}{CK}$

Mà $\widehat{KCI}=\widehat{ECD}$

$\to\Delta CIK\sim\Delta CDE(c.g.c)$

$\to \widehat{CIK}=\widehat{CDE}$

$\to 180^o-\widehat{CIK}=180^o-\widehat{CDE}$

$\to \widehat{KIE}=\widehat{ADE}$

Mặt khác $\widehat{KEI}=\widehat{EKC}=\widehat{IDC}=\widehat{EAD}$

$\to\Delta KEI\sim\Delta EAD(g.g)$

Dễ chứng minh $\Delta EAD\sim\Delta CAB(c.g.c)$

$\to \Delta KEI\sim\Delta CAB$

$\to \dfrac{S_{KEI}}{S_{CAB}}=(\dfrac{KE}{CA})^2$

Xét $\Delta AKE,\Delta KEC$ có:

$\widehat{AKE}=\widehat{CKE}(=90^o)$

$\widehat{AEK}=90^o-\widehat{KEC}=\widehat{KCE}$

$\to\Delta AKE\sim\Delta EKC(g.g)$

$\to \dfrac{AK}{KE}=\dfrac{EK}{KC}$

$\to EK^2=KA.KC\le\dfrac14(KA+KC)^2=\dfrac14AC^2$

$\to \dfrac{EK^2}{AC^2}\le\dfrac14$

$\to \dfrac{S_{KEI}}{S_{CAB}}\le\dfrac14$

$\to S_{ABC}\ge 4S_{EIK}$

Đáp án:

Giải thích các bước giải: