Giải thích các bước giải:
a.Xét $\Delta HBA , \Delta ABC$ có:
Chung $\hat B$
$\widehat{AHB}=\widehat{BAC}(=90^o)$
$\to\Delta HBA\sim\Delta ABC(g.g)$
$\to \dfrac{BA}{BC}=\dfrac{HB}{AB}$
$\to AB^2=BH.BC$
b.Xét $\Delta AHB,\Delta HAC$ có:
$\widehat{AHB}=\widehat{AHC}(=90^o)$
$\widehat{BAH}=90^o-\widehat{HAC}=\widehat{ACH}$
$\to\Delta HBA\sim\Delta HAC(g.g)$
$\to\dfrac{HB}{HA}=\dfrac{HA}{HC}$
$\to HA^2=HB.HC$
c.Ta có $HD\perp AC, AB\perp AC\to HD//AB$
$\to \dfrac{EH}{MB}=\dfrac{CE}{CM}=\dfrac{DE}{AM}$
$\to EH=ED$ vì $MB=MA$ do $M$ là trung điểm $AB$
Mà $AB//DH$
$\to \widehat{IBM}=\widehat{IDE}$
Mặt khác $\dfrac{BM}{DE}=\dfrac{2BM}{2DE}=\dfrac{AB}{DH}=\dfrac{BI}{ID}$
$\Delta BMI\sim\Delta DEI(c.g.c)$
$\to \widehat{MIB}=\widehat{DIE}$
$\to B,I,D$ thẳng hàng