Quadratic progression?

For the given equation to be a perfect square, its discriminant should be $0$ . $D=0$ $b^{2}(c-a)^{2}=4ac(a-b)(b-c)$ Divide both sides by $a^{2}b^{2}c^{2}$ . $\Bigg( \dfrac{1}{a} - \dfrac{1}{c}\Bigg)^{2} = 4\Bigg( \dfrac{1}{b}-\dfrac{1}{a}\Bigg) \Bigg( \dfrac{1}{c}-\dfrac{1}{b}\Bigg)$ After expanding the brackets and taking all the terms on $LHS$ we get:- $\dfrac{1}{a^{2}}+ \dfrac{4}{b^{2}} + \dfrac{1}{c^{2}}+\dfrac{2}{ac}-\dfrac{4}{ab}-\dfrac{4}{bc} = 0$ $\Bigg( \dfrac{1}{a}-\dfrac{2}{b}+\dfrac{1}{c}\Bigg)^{2}=0$ $\implies \dfrac{2}{b} = \dfrac{1}{c} + \dfrac{1}{a}$ Hence $a,b,c$ are in $HP$ .

Since it is a perfect square therefore Discriminant = 0 . By solving the discriminant we get, $b = \frac{2ab}{a+b}$ and hence $a, b, c$ are in H.P

Considering it to be pair of straight lines we use formula for angle between them h^2=ab and deduce from there.