c, Sai đề, đáng lẽ là $AH.BC$ nhé
d, Xét $ΔAHB$ và $ΔCAB$ có:
$\widehat{AHB}=\widehat{CAB}=90^o$
$\widehat{B}$ chung
$⇒ΔAHB \sim ΔCAB(g.g)$
$⇒\dfrac{AH}{AC}=\widehat{AB}{BC}$
$⇒( \dfrac{AH}{AC})^2=(\dfrac{AB}{BC})^2$
Chứng minh tương tự ta có:
$ΔAHC \sim ΔBAC(g.g)$
$⇒\dfrac{AH}{AB}=\dfrac{AC}{BC}$
$⇒(\dfrac{AH}{AB})^2=(\dfrac{AC}{BC})^2$
$⇒( \dfrac{AH}{AC})^2+(\dfrac{AH}{AB})^2=(\dfrac{AB}{BC})^2+(\dfrac{AC}{BC})^2$
Hay $\dfrac{AH^2}{AC^2}+\dfrac{AH^2}{AB^2}=\dfrac{AB^2+AC^2}{BC^2}=\dfrac{BC^2}{BC^2}=1$
Chia cả 2 vế cho $AH^2$ nên ta có:
$\dfrac{1}{AC^2}+\dfrac{1}{AB^2}=\dfrac{1}{AH^2}$ đpcm
