a/ Xét $ΔBHA$ và $ΔBAC$:
$\widehat B:chung$
$\widehat{BHA}=\widehat{BAC}(=90^\circ)$
$→ΔBHA\backsim ΔBAC(g-g)$
Xét $ΔCHA$ và $ΔCAB$:
$\widehat C:chung$
$\widehat{CHA}=\widehat{CAB}(=90^\circ)$
$→ΔCHA\backsim ΔCAB(g-g)$
b/ $ΔBHA\backsim ΔBAC$
$→\dfrac{BH}{BA}=\dfrac{BA}{BC}$
$↔BA^2=BH.BC$
$ΔCHA\backsim ΔCAB$
$→\dfrac{CH}{CA}=\dfrac{CA}{CB}$
$↔CA^2=CH.CB$
c/ $ΔBHA\backsim ΔBAC$ hay $ΔAHB\backsim ΔCAB$
mà $ΔCHA\backsim ΔCAB$
$→ΔAHB\backsim ΔCHA$
$→\dfrac{AH}{HB}=\dfrac{CH}{AH}$
$↔AH^2=HB.CH$
d/ Xét $ΔAEH$ và $ΔAHB$
$\widehat A:chung$
$\widehat{AEH}=\widehat{AHB}(=90^\circ)$
$→ΔAEH\backsim ΔAHB(g-g)$
$→\dfrac{AE}{AH}=\dfrac{AH}{AB}$
$↔AH^2=AE.AB$
Xét $ΔAFH$ và $ΔAHC$:
$\widehat A:chung$
$\widehat{AFH}=\widehat{AHC}(=90^\circ)$
$→ΔAFH\backsimΔ AHC(g-g)$
$→\dfrac{AF}{AH}=\dfrac{AH}{AC}$
$↔AH^2=AF.AC$
Ta có: $\begin{cases}AH^2=AE.AB\\AH^2=AF.AC\end{cases}$
$→AE.AB=AF.AC$
e/ $AE.AB=AF.AC$
$↔\dfrac{AE}{AF}=\dfrac{AC}{AB}$
Xét $ΔAEF$ và $ΔACB$:
$\dfrac{AE}{AF}=\dfrac{AC}{AB}(cmt)$
$\widehat A:chung$
$→ΔAEF\backsim ΔACB(c-g-c)$
