Giải thích các bước giải:

a.Xét $\Delta BHA,\Delta ABC$ có:

Chung $\hat B$

$\widehat{BHA}=\widehat{BAC}(=90^o)$

$\to\Delta BHA\sim\Delta BAC(g.g)$

b.Xét $\Delta AHB,\Delta AHC$ có:

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

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

$\to \Delta HAB\sim\Delta HCA(g.g)$

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

$\to AH^2=HB.HC$

c.Vì $P$ là trung điểm $AB$

$\to PA=PB$

Ta có: $HM//AB(\perp AC)$

$\to \dfrac{QH}{PB}=\dfrac{CQ}{CP}=\dfrac{QM}{AP}$

$\to QH=QM$

$\to Q$ là trung điểm $HM$

$\to \dfrac{QM}{PB}=\dfrac{QH}{AP}=\dfrac{IQ}{IP}$

Mà $\widehat{IQM}=\widehat{IPB}$ vì $HM//AB$

$\to \Delta IQM\sim\Delta IPB(c.g.c)$

$\to \widehat{MIQ}=\widehat{PIB}$

$\to \widehat{BIM}=\widehat{BIP}+\widehat{PIM}=\widehat{MIQ}+\widehat{PIM}=\widehat{PIQ}=180^o$

$\to B, I, M$ thẳng hàng