Xét $ΔBAD$ và $ΔCAB$ có:
$\widehat{ABD}=\widehat{ACB}$ (do $\widehat{ABD}=\widehat{C}$)
$\widehat{A}$ chung
$⇒ΔBAD$$\sim$$ΔCAB(g.g)$
$⇒\dfrac{AB}{AC}=\dfrac{BD}{CB}$
Hay $\dfrac{AB}{BD}=\dfrac{AC}{CB}$
$⇒(\dfrac{AB}{BD})^2=(\dfrac{AC}{CB})^2$
$⇒(\dfrac{AB}{BD})^2+(\dfrac{AB}{BC})^2=(\dfrac{AC}{CB})^2+(\dfrac{AB}{BC})^2=\dfrac{AB^2+AC^2}{BC^2}=\dfrac{BC^2}{BC^2}=1$
(do $ΔABC$ vuông tại $A$⇒$AB^2+AC^2=BC^2$(định lí Pitago))
$⇒(\dfrac{AB}{BD})^2+(\dfrac{AB}{BC})^2=1$
$⇔\dfrac{AB^2}{BD^2}+\dfrac{AB^2}{BC^2}=\dfrac{AB^2}{AB^2}$
$⇔AB^2.\dfrac{1}{BD^2}+AB^2.\dfrac{1}{BC^2}=AB^2\dfrac{1}{AB^2}$
$⇔\dfrac{1}{BD^2}+\dfrac{1}{BC^2}=\dfrac{1}{AB^2}$
$⇒đpcm$
