Giải thích các bước giải:
Ta có:
$\dfrac{a+b}{5}=\dfrac{b+c}{7}=\dfrac{a+c}{8}=\dfrac{(a+b)+(b+c)+(a+c)}{5+7+8}=\dfrac{2(a+b+c)}{20}=\dfrac{a+b+c}{10}$
$\to \dfrac{a+b+c-(a+b)}{10-5}=\dfrac{(a+b+c)-(b+c)}{10-7}=\dfrac{a+b+c-(a+c)}{10-8}$
$\to \dfrac{c}{5}=\dfrac{a}{3}=\dfrac{b}{2}=k$
$\to a =3k, b=2k, c=5k$
Mà $S_{ABC}=\dfrac12ah_a=\dfrac12bh_b=\dfrac12ch_c$
$\to ah_a=bh_b=ch_c$
$\to 3kh_a=2kh_b=5kh_c$
$\to 3h_a=2h_b=5h_c$
$\to \dfrac{3h_a}{30}=\dfrac{2h_b}{30}=\dfrac{5h_c}{30}$
$\to \dfrac{h_a}{10}=\dfrac{h_b}{15}=\dfrac{h_c}{6}$