Giải thích các bước giải:


a.Xét $\Delta AHB,\Delta AHC$ có:

$\widehat{AHB}=\widehat{AHC}(=90^o)$

$\widehat{HAB}=90^o-\hat B=\hat C$

$\to \Delta AHB\sim\Delta CHA(g.g)$

$\to \dfrac{HA}{HC}=\dfrac{HB}{HA}$

$\to AH^2=HB.HC$

b.Vì $DE//AC(\perp AB), DK//AB(\perp AC)$

        $DE\perp AB, DK\perp AC, AB\perp AC, AD$ là phân giác $\hat A\to AEDK$ là hình vuông $\to AE=DE=DK=KA$

$\to \dfrac{ME}{MD}=\dfrac{KA}{KC}=\dfrac{DE}{KC}=\dfrac{NE}{NC}$

c.Ta có: $DK//AB$

$\to\dfrac{KN}{AE}=\dfrac{CK}{CA}=\dfrac{DK}{AB}$

$\to \dfrac{KN}{AK}=\dfrac{KA}{AB}$

Mà $\widehat{BAK}=\widehat{AKN}(=90^o)$

$\to \Delta ABK\sim\Delta KAN(c.g.c)$

Gọi $AN\cap BK=G$

$\to \widehat{NAK}=\widehat{ABK}\to \widehat{GKA}=\widehat{ABK}$

$\to \Delta KGA\sim\Delta KAB(g.g)$

$\to \widehat{KGA}=\widehat{KAB}=90^o$

$\to AN\perp BK$

$\to MI\perp AN$

Tương tự $NI\perp AM$

$\to I$ là trực tâm $\Delta AMN$

$\to AI\perp MN$

Mà $MN //BC\to AI\perp BC$

$\to A, I, H$ thẳng hàng

d.Ta có:

$\dfrac{MP}{PA}=\dfrac{S_{MBD}}{S_{ABD}}$
$\dfrac{ME}{DE}=\dfrac{S_{MAB}}{S_{ABD}}$

$\dfrac{MF}{FB}=\dfrac{S_{MAD}}{S_{ABD}}$

Cộng vế với vế

$\to \dfrac{MP}{AP}+\dfrac{ME}{DE}+\dfrac{MF}{FB}=\dfrac{S_{MBD}+S_{MAB}+S_{AMD}}{S_{ABD}}=1$

$\to \dfrac{AP}{MP}+\dfrac{BF}{MF}+\dfrac{DE}{ME}=\dfrac{9}{\dfrac{MP}{AP}}+\dfrac{9}{\dfrac{ME}{DE}}+\dfrac{9}{\dfrac{FM}{FB}}$

$\to \dfrac{AP}{MP}+\dfrac{BF}{MF}+\dfrac{DE}{ME}\ge\dfrac9{\dfrac{MP}{AP}+\dfrac{ME}{DE}+\dfrac{MF}{FB}}$

$\to \dfrac{AP}{MP}+\dfrac{BF}{MF}+\dfrac{DE}{ME}\ge 9$

$\to đpcm$