a/ Xét $ΔBHA$ và $ΔBAC$:
$\widehat B:chung$
$\widehat{BHA}=\widehat{BAC}(=90^\circ)$
$→ΔBHA\backsim ΔBAC(g-g)$
$→\dfrac{BH}{BA}=\dfrac{BA}{BC}$
$↔BA^2=BH.BC$ hay $AB^2=BC.BH$
Xét $ΔAHC$ và $ΔBAC$:
$\widehat C:chung$
$\widehat{AHC}=\widehat{BAC}(=90^\circ)$
$→ΔAHC\backsim ΔBAC(g-g)$
$→\dfrac{CH}{CA}=\dfrac{CA}{CB}$
$↔CA^2=CH.CB$ hay $AC^2=BC.CH$
Vậy $AB^2=BC.CH$ và $AC^2=BC.CH$
b/ Ta có: $\begin{cases}ΔBHA\backsim ΔBAC(cmt)\\ΔAHC\backsim ΔBAC(cmt)\end{cases}$
$→ΔBHA\backsim ΔAHC$
$→\dfrac{BH}{AH}=\dfrac{AH}{CH}$
$↔AH^2=BH.CH$
Vậy $AH^2=BH.CH$
c/ Vì $ΔABC$ vuông tại $A$
$→S_{ΔABC}=\dfrac{1}{2}.AB.AC$
mà $S_{ΔABC}=\dfrac{1}{2}.AH.BC$
$→\dfrac{1}{2}.AB.AC=\dfrac{1}{2}.AH.BC\\↔AB.AC=AH.BC$
Vậy $AB.AC=AH.BC$
