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 $ΔACB$:

$\dfrac{AE}{AF}=\dfrac{AC}{AB}(cmt)$

$\widehat A:chung$

$→ΔAEF\backsim ΔACB(c-g-c)$