For $\triangle ABC$ with inradius $r$, if $\frac{c + a}{b} +\frac{c + b}{a} = \frac{c}{r}$, then which angle is $90^\circ$?

Question

For $\triangle ABC$ with inradius $r$, if $$\frac{c + a}{b} +\frac{c + b}{a} = \frac{c}{r}$$ then which angle is $90^\circ$?

My try :

I am stuck here I think my process is wrong please help me in this.

Answer 1

The equality on the $3^{rd}$ line can be written as:

$$r = \frac{abc}{a^2+b^2+(a+b)c}$$

The symmetry in $\,a,b\,$ suggests that the right angle is $\,C\,$, in which case $\,a^2+b^2=c^2\,$, then:

$$\require{cancel} r = \frac{abc}{c^2+(a+b)c} = \frac{ab\cancel{c}}{(a+b+c)\cancel{c}} =\frac{ab}{a+b+c} = \frac{\cancel{2}S}{\cancel{2}p} = \dfrac{S}{p} \tag{1} $$

In any triangle the inradius $\,r = \dfrac{S}{p}\,$ where $\,S\,$ is the area and $\,p = \dfrac{a+b+c}{2}\,$ is the semi-perimeter. For a right triangle $S = \dfrac{1}{2}ab\,$, so $(1)$ holds true, therefore the right angle is indeed $\,C\,$.