$\text{Áp dụng hệ thức lượng trong $∆ABC$ có đường cao AH}$
$a)AH^2=HB.HC⇒HB=\dfrac{AH^2}{HC}=\dfrac{4^2}{6}=\dfrac{8}{3}cm$
$BC=HB+HC=\dfrac{8}{3}+6=\dfrac{26}{3}cm$
$AB^2=HB.BC⇒AB=\sqrt{HB.BC}=\sqrt{\dfrac{8}{3}.\dfrac{26}{3}}=\dfrac{4\sqrt{13}}{3}cm$
$AC^2=HC.BC⇒AC=\sqrt{HC.BC}=\sqrt{6.\dfrac{26}{3}}=2\sqrt{13}cm$
$b)AH^2=HB.HC⇒AH=\sqrt{HB.HC}=\sqrt{2.4}=\sqrt{6}cm$
$BC=HB+HC=2+4=6cm$
$AB=\sqrt{HB.BC}=\sqrt{2.6}=2\sqrt{3}cm$
$AC=\sqrt{HC.BC}=\sqrt{4.6}=2\sqrt{6}cm$
$c)$ Áp dụng định lý Pytago trong $∆ABC$ vuông tại $\hat{A}$
$AB^2+AC^2=BC^2$
$⇒AB=\sqrt{BC^2-AC^2}=\sqrt{13^2-12^2}=5cm$
$AH.BC=AB.AC⇒AH=\dfrac{AB.AC}{BC}=\dfrac{5.12}{13}=\dfrac{60}{13}cm$
$AB^2=HB.BC⇒HB=\dfrac{AB^2}{BC}=\dfrac{5^2}{13}=\dfrac{25}{13}cm$
