a/ Xét $ΔBHA$ và $ΔBAC$:
$\widehat{ABH}\,\,hay\,\,\widehat{CBA}:chung$
$\widehat{BHA}=\widehat{BAC}(=90^\circ)$
$→ΔBHA\backsim ΔBAC(g-g)$
$→\dfrac{BH}{BA}=\dfrac{BA}{BC}$
$↔AB^2=BH.BC$
Xét $ΔBAC$ và $ΔAHC$:
$\widehat{BCA}\,\,hay\,\,\widehat{ACH}:chung$
$\widehat{BAC}=\widehat{AHC}(=90^\circ)$
$→ΔBAC\backsim ΔAHC(g-g)$
$→\dfrac{AC}{BC}=\dfrac{HC}{AC}$
$↔AC^2=CH.BC$
b/ $\begin{cases}ΔBHA\backsim ΔBAC\\ΔBAC\backsim ΔAHC\end{cases}\\→ΔBHA\backsim ΔAHC\\→\dfrac{AH}{BH}=\dfrac{CH}{AH}\\↔AH^2=BH.CH$
c/ $ΔBHA\backsim ΔBAC$
$→\dfrac{AB}{AH}=\dfrac{CB}{CA}$
$↔AB.AC=AH.BC$
d/ $\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\\=\dfrac{AB^2+AC^2}{(AB.AC)^2}\\=\dfrac{BH.BC+CH.BC}{(AH.BC)^2}\\=\dfrac{BC(BH+CH)}{AH^2.BC^2}\\=\dfrac{BC.BC}{AH^2.BC^2}\\=\dfrac{BC^2}{AH^2.BC^2}\\=\dfrac{1}{AH^2}$
