a/ Xét $ΔABH$ và $ΔCBA$:
$\widehat B:chung$
$\widehat{AHB}=\widehat{CAB}(=90^\circ)$
$→ΔABH\backsim ΔCBA(g-g)$
b/ $ΔABH\backsim ΔCBA$
$→\dfrac{AB}{BH}=\dfrac{CB}{AB}$
$↔AB^2=BH.CB$ hay $AB^2=4.13$
$↔AB^2=52\\↔AB=2\sqrt{13}cm$
c/ Vì $\widehat{EHA},\widehat{FHC}$ cùng phụ $\widehat{AHF}$
$→\widehat{EHA}=\widehat{FHC}$
$ΔABH\backsim ΔCBA$
$→\widehat{BAH}=\widehat{BCA}$ hay $\widehat{EAH}=\widehat{FCH}$
Xét $ΔEAH$ và $ΔFCH$:
$\widehat{EHA}=\widehat{FHC}(cmt)$
$\widehat{EAH}=\widehat{FCH}(cmt)$
$→ΔEAH\backsim ΔFCH(g-g)$
$→\dfrac{AE}{AH}=\dfrac{CF}{CH}$
$↔AE.CH=AH.CF$