a/ Xét $ΔBHA$ và $ΔBAC$:
$\widehat B:chung$
$\widehat{BHA}=\widehat{BAC}(=90^\circ)$
$→ΔBHA\backsim ΔBAC(g-g)$
$→\dfrac{AH}{AB}=\dfrac{AC}{BC}$
$↔AH.BC=AB.AC$
b/ $ΔBHA\backsim ΔBAC$
$→\dfrac{BH}{AB}=\dfrac{AB}{BC}$
$↔AB^2=BH.BC$ hay $AB^2=4.13$
$↔AB^2=52$
$↔AB=2\sqrt{13}cm$
c/ $ΔBHA\backsim ΔBAC$
$→\widehat{BAH}=\widehat{BCA}$
hay $\widehat{HAE}=\widehat{HCF}$
Ta có: $\widehat{EHA}$ và $\widehat{FHC}$ cùng phụ $\widehat{AHF}$
$→\widehat{EHA}=\widehat{FHC}$
Xét $ΔEHA$ và $ΔFHC$:
$\widehat{EHA}=\widehat{FHC}(cmt)$
$\widehat{HAE}=\widehat{HCF}(cmt)$
$→ΔEHA\backsim ΔFHC(g-g)$
$→\dfrac{AH}{AE}=\dfrac{CH}{CF}$
$↔AE.CH=AH.CF$ hay $AE.CH=AH.FC$
