In a triangle $ABC$, if an exradius is the sum of the two other exradii and the inradius, the triangle is
- equilateral
- isosceles
- scalene
- right-angled
I've tried multiple times, but this is the most simplified expression I've got: $$r_1 = r_2 + r_3 + r$$ $$sX\tan\frac A2=sX\tan\frac B2+sX\tan\frac C2+(s-a)\tan\frac A2$$ $$sX\tan\frac A2=sX\tan\frac B2+sX\tan\frac C2+sX\tan\frac A2-aX\tan\frac A2$$ $$aX\tan\frac A2=sX\tan\frac B2+sX\tan\frac C2$$ I could not go further.