<s>[INST] A problem relating to triangles and progressions If $D,E,F$ are the points of contact of the inscribed circle with the sides $BC, CA, AB$ of a triangle $\triangle ABC$, we need to show that if the squares of $AD, BE, CF$ are in arithmetic progression, then the sides of a triangle are in harmonic progression.
I tried using cosine law:
$$AD\times AD = c^2 +(s-b)^2 -2c(s-b)\cos B\\
BE\times BE = a^2 + (s-c)^2 -2a(s-c)\cos C\\
CF\times CF = b^2 + (s-a)^2 -2b(s-a)\cos A\\$$
But I couldn't simplify further to prove that the sides of a triangle are in harmonic progression.
 [/INST] $$AD^2=c^2+(s-b)^2-2c(s-b)cosB=(c-(s-b))^2+2c(s-b)(1-cosB)=(b+c-s)^2+4c(s-b)sin^2(\frac{B}{2})$$  But $$sin^2\left(\frac{B}{2}\right)=\frac{(s-a)(s-c)}{ac}$$ So
$$AD^2=(b+c-s)^2+4c(s-b)\frac{(s-a)(s-c)}{ac}=(s-a)^2+4\frac{\Delta^2}{as}=(s-a)^2+\frac{4r\Delta}{a}$$  Similarly
$$BE^2=(s-b)^2+\frac{4r\Delta}{b}$$ and
$$CF^2=(s-c)^2+\frac{4r\Delta}{c}$$  Now since $AD^2$,$BE^2$ and $CF^2$ are in A.P we have
$$BE^2-AD^2=CF^2-BE^2$$ $\implies$
$$(s-b)^2-(s-a)^2+4r\Delta\left(\frac{1}{b}-\frac{1}{a}\right)=(s-c)^2-(s-b)^2+4r\Delta\left(\frac{1}{c}-\frac{1}{b}\right)$$ $\implies$
$$(a-b)c+4r\Delta\left(\frac{1}{b}-\frac{1}{a}\right)=(b-c)a+4r\Delta\left(\frac{1}{c}-\frac{1}{b}\right)$$ $\implies$
$$2ac-b(a+c)=4r\Delta\left(\frac{1}{a}+\frac{1}{c}-\frac{2}{b}\right)$$ from that
$$2ac-b(a+c)=4r\Delta\left(\frac{(a+c)b-2ac}{abc}\right)$$ which is possible only if
$$2ac=b(a+c)$$ so $a$,$b$ and $c$ are in H.P
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