A condition under which sidelengths of a triangle are in HP

Question

If D , E , F are points of contact of the inscribed circle with sides BC , CA , AB of a triangle , show that if the squares of AD , BE , CF are in arithmetic progression, then the sides of the triangle are in harmonic progression I found this question in SL Loney - Plane Trigonmetry

Answer 1

By the given and by law of cosines we have $$AD^2+CF^2=2BE^2$$ or $$c^2+\frac{(a+c-b)^2}{4}-\frac{(a+c-b)(a^2+c^2-b^2)}{2a}+$$ $$+a^2+\frac{(a+c-b)^2}{4}-\frac{(a+c-b)(a^2+c^2-b^2)}{2c}=$$ $$=2\left(c^2+\frac{(b+c-a)^2}{4}-\frac{(b+c-a)(b^2+c^2-a^2)}{2b}\right)$$ or $$(2ac-ab-bc)\sum_{cyc}a^2(b+c-a)=0$$ or $$ab+bc=2ac$$ or $$\frac{\frac{1}{a}+\frac{1}{c}}{2}=\frac{1}{b}.$$ Done!

Answer 2

It is well known that the touching points joined to their opposite points concur at the Gergonne point. (But we do not need this).

It is easy to show $DC=\frac{a+b-c}{2}$ and after a little trigonometry and algebra we have \begin{eqnarray*} x^2= \frac{(-a+b+c)(a^2+a(b+c)-2(b-c)^2)}{4a} \\ y^2= \frac{(a-b+c)(b^2+b(a+c)-2(a-c)^2)}{4b} \\ z^2= \frac{(a+b-c)(c^2+c(b+a)-2(b-a)^2)}{4c} \\ \end{eqnarray*} Now for these to be in Arithmetic progression we require that the sum of two of them equals $2$ times the third ... lets see what reduce makes of this ?

So we have $2ab-bc-ca=0$ , in other words the sides are in harmonic progression. (Note that the other bracket cannot be zero).