Xét $ΔBHA$ và $ΔBAC$:
$\widehat B:chung$
$\widehat{BHA}=\widehat{BAC}(=90^\circ)$
$→ΔBHA\backsim ΔBAC(g-g)$
$→\dfrac{BH}{AB}=\dfrac{AB}{BC}$
$↔AB^2=BH.BC$
Xét $ΔCHA$ và $ΔCAB$:
$\widehat C:chung$
$\widehat{CHA}=\widehat{CAB}(=90^\circ)$
$→ΔCHA\backsim ΔCAB(g-g)$
$→\dfrac{CH}{AC}=\dfrac{AC}{BC}$
$↔AC^2=CH.BC$
$\dfrac{HC}{HB}=\dfrac{HC.BC}{HB.BC}=\dfrac{AC^2}{AB^2}=\bigg(\dfrac{AC}{AB}\bigg)^2\\→\dfrac{HC}{HB}=\bigg(\dfrac{3}{4}\bigg)^2=\dfrac{9}{16}$
Vậy $\dfrac{HC}{HB}=\dfrac{9}{16}$