Giải thích các bước giải:
a.Ta có $\Delta ABC$ vuông tại $A, M$ là trung điểm $BC\to AM=MB=MC=\dfrac12BC$
Ta có: $\widehat{BHA}=\widehat{BAC}=90^o,\widehat{ABH}=\widehat{ABC}$
$\to\Delta BAH\sim\Delta BCA(g.g)$
$\to\dfrac{BA}{BC}=\dfrac{BH}{BA}$
$\to BA^2=BH.BC=BH.2AM$
$\to AB^2=2BH.AM$
b.Ta có:
$\widehat{BEA}=\widehat{BAF}=90^o,\widehat{ABE}=\widehat{ABF}$
$\to\Delta BAE\sim\Delta BFA(g.g)$
$\to\dfrac{BA}{BF}=\dfrac{BE}{BA}$
$\to BA^2=BE.BF$
$\to BE.BF=BH.BC$
Ta có:
$\widehat{ABF}=\widehat{ABE}=90^o-\widehat{BAE}=\widehat{EAF}=\widehat{MAC}=\widehat{MCA}=\widehat{ACB}$
$\widehat{BAF}=\widehat{BAC}$
$\to\Delta ABF\sim\Delta ACB(g.g)$
$\to\dfrac{AB}{AC}=\dfrac{AF}{AB}$
$\to AB^2=AC.AF$
$\to BE.BF=BH.BC=AF.AC$
