Giải thích các bước giải:
a.Xét $\Delta ABD,\Delta BHI$ có:
$\widehat{ABD}=\widehat{HBI}$ vì $BD$ là phân giác $\hat B$
$\widehat{BAD}=\widehat{BHI}(=90^o)$
$\to\Delta ABD\sim\Delta HBI(g.g)$
b.Xét $\Delta ABH,\Delta ABC$ có:
Chung $\hat B$
$\widehat{AHB}=\widehat{BAC}(=90^o)$
$\to\Delta BAH\sim\Delta BCA(g.g)$
$\to\dfrac{BA}{BC}=\dfrac{BH}{BA}$
$\to AB^2=BH.BC$
Xét $\Delta AHB,\Delta AHC$ có:
$\widehat{AHB}=\widehat{AHC}(=90^o)$
$\widehat{BAH}=90^o-\widehat{HAC}=\widehat{HCA}$
$\to\Delta HAB\sim\Delta HCA(g.g)$
$\to\dfrac{HA}{HC}=\dfrac{HB}{HA}$
$\to AH^2=HB.HC=144$
$\to AH=12$
c.Từ câu a
$\to \widehat{ADB}=\widehat{BIH}$
$\to\widehat{ADI}=\widehat{AID}$
$\to\Delta ADI$ cân tại $A$
$\to IA=AD$
Ta có $BD$ là phân giác $\hat B$
$\to \dfrac{IA}{IH}=\dfrac{BA}{BH}=\dfrac{BC}{BA}=\dfrac{DC}{DA}$
$\to IA.DA=IH.DC$
Mà $IA=AD$
$\to AD^2=IH.DC$
d.Ta có $KP\perp AC, KC\perp DB$
$\to \widehat{PKC}=90^o-\widehat{KCP}=90^o-\widehat{KCD}=\widehat{KDC}=\widehat{ADB}=90^o-\widehat{ABD}=90^o-\dfrac12\hat B$
Vì $KB\perp KC, Q$ là trung điểm $BC$
$\to QK=QB=QC$
$\to\Delta KQC,\Delta KQB$ cân tại $Q$
$\to\widehat{QKC}=\widehat{KCQ}=\widehat{KCB}=90^o-\widehat{KBC}=90^o-\dfrac12\hat B$
$\to \widehat{PKC}=\widehat{QKC}$
$\to K, P, Q$ thẳng hàng
