Call $ E,F$ the points respectively on $ CA,AB$ such that $ BE,CF$ bisect the perimeter, and call $ s$ the semiperimeter. Clearly $ CE = s - a$, $ AE = s - c$, and by Stewart, \[ BE^2 = \frac {(s - a)c^2 + (s - c)a^2}{b} - (s - a)(s - c), \] and similarly for $ CF$. Therefore, \[ BE = CF\leftrightarrow\frac {(s - a)c^2 + (s - c)a^2}{b} - (s - a)(s - c) = \frac {(s - a)b^2 + (s - b)a^2}{c} - (s - a)(s - b)\leftrightarrow(b - c)(s - a)^2(a + b + c) = 0. \] Clearly $ a + b + c > 0$ and by the triangular inequality $ s - a > 0$. Hence two such cevians as described in the problem are equal iff the corresponding sides are equal. It follows that the answer is yes, the lengths are necesarily different.

No computations! In the notations of the previous post, $ BC + CE = CB + BF = \frac {1} {2} (AB + BC + CA)$, hence $ BF = CE$. Then, if $ BE = CF$, triangles $ \Delta BCF$ and $ \Delta CBE$ are congruent, hence $ \angle BCE = \angle CBF$, so $ \Delta ABC$ is isosceles.