Giải thích các bước giải:
a.Xét $\Delta ABD, \Delta ACE$ có:
Chung $\hat A$
$\widehat{ADB}=\widehat{AEC}(=90^o)$
$\to \Delta ABD\sim\Delta ACE(g.g)$
$\to \dfrac{AB}{AC}=\dfrac{AD}{AE}$
$\to AE.AB =AD.AC$
b.Xét $\Delta HDC, \Delta HEB$ có:
$\widehat{HDC}=\widehat{HEB}(=90^o)$
$\widehat{DHC}=\widehat{EHB}$
$\to \Delta HDC\sim\Delta HEB(g.g)$
$\to \dfrac{HD}{HE}=\dfrac{HC}{HB}$
$\to HD.HB=HC.HE$
c.Ta có $BD\perp AC, CE\perp AB, BD\cap CE=H$
$\to H$ là trực tâm $\Delta ABC\to AH\perp BC$
Mà $HM\perp BC$
$\to A, H, M$ thẳng hàng
Xét $\Delta HBM, \Delta AMC$ có:
$\widehat{HMB}=\widehat{AMC}(=90^o)$
$\widehat{BHM}=90^o-\widehat{HBM}=90^o-\widehat{DBC}=\widehat{DCB}=\widehat{ACM}$
$\to \Delta MBH\sim\Delta MAC(g.g)$
$\to \dfrac{MB}{MA}=\dfrac{MH}{MC}$
$\to MB.MC=MH.MA$
d.Ta có $\hat C=30^o, AM\perp BC$
$\to \Delta AMC$ là nửa tam giác đều
$\to AM=\dfrac12AC$
Xét $\Delta ADE, \Delta ABC$ có:
Chung $\hat A$
$\dfrac{AD}{AB}=\dfrac{AE}{AC}$ vì $AD.AC=AE.AB$
$\to \Delta ADE\sim\Delta ABC(c.g.c)$
$\to \dfrac{S_{ADE}}{S_{ABC}}=(\dfrac{AE}{AC})^2$
$\to \dfrac{S_{ABC}-S_{ADE}}{S_{ABC}}=\dfrac{AC^2-AE^2}{AC^2}$
$\to \dfrac{S_{BCDE}}{S_{ABC}}=\dfrac{AC^2-AE^2}{AC^2}$
$\to S_{BEDC}=\dfrac{AC^2-AE^2}{AC^2}\cdot S_{ABC}$
$\to S_{BEDC}=\dfrac{AC^2-AE^2}{AC^2}\cdot 400$
Xem lại đề câu d
