(a,b,c) in H.P. ⇒ (\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}) in A.P. ⇒ (2/b=1/a+1/c) ⇒ (b=\dfrac{2ac}{a+c})

(i)

[ \dfrac{a-b}{b-c}=\dfrac{a-\frac{2ac}{a+c}}{\frac{2ac}{a+c}-c}=\dfrac{\frac{a(a+c)-2ac}{a+c}}{\frac{2ac-c(a+c)}{a+c}}=\dfrac{a^2-ac}{2ac-ac-c^2}=\dfrac{a(a-c)}{c(a-c)}=\dfrac{a}{c} ]

(ii)

[ (a-c)^2=(a+c)(a-2b+c) ] follows by substituting (b=\dfrac{2ac}{a+c}).

Proved.