Let the angle bisectors be $AD$ and $BE$. Then \[[ABC] = \frac{1}{2} \cdot BC \cdot AD \cdot \sin \angle ADB\] and \[[ABC] = \frac{1}{2} \cdot AC \cdot BE \cdot \sin \angle BEA.\] Since $AD \cdot BC = BE \cdot AC$, $\sin \angle ADB = \sin \angle BEA$, or \[\sin \left( \frac{A}{2} + C \right) = \sin \left( \frac{B}{2} + C \right),\] which means \[\sin \left( \frac{A}{2} + C \right) - \sin \left( \frac{B}{2} + C \right) = 2 \sin \frac{A - B}{4} \cos \frac{A + B + 4C}{4} = 0.\] If \[\sin \frac{A - B}{4} = 0,\] then $A = B$. If \[\cos \frac{A + B + 4C}{4} = 0,\] then $(A + B + 4C)/4 = (3C + 180^\circ)/4 = 90^\circ$, so $C = 60^\circ$.

Let $N$ and $M$ be the feet of altitudes from $A$ AND $B$, respectively. We know that $AD\cdot BC = BE\cdot AC$, and that $AN\cdot BC = BM\cdot AC$, so $\frac{AD}{AN} = \frac{BE}{BM}$. Therefore, $\angle MEB = \angle NDA$. We have two cases: Case 1: AM-AE and BD-BN have the same sign (they won't be 0; otherwise it would be isosceles) Because $\angle MEB = \angle NDA$, then $A$, $E$, $D$, and $B$ are concyclic. Because $AD$ and $BE$ are angle bisectors, $\angle BAD = \angle EAD = \angle EBD = \angle EBA$, so $\angle A = \angle B$, a contradiction to the non-isosceles hypothesis. Case 2: AM-AE and BD-BN have different signs. [asy][asy] import graph; size(10.11cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5.88, xmax = 21.23, ymin = -5.67, ymax = 8.62; /* image dimensions */ pen ttwwff = rgb(0.2,0.4,1); draw((5.29,1.56)--(5.66,1.86)--(5.35,2.23)--(4.99,1.92)--cycle, ttwwff); draw((9.24,-1.88)--(9.3,-2.23)--(9.65,-2.16)--(9.59,-1.81)--cycle, ttwwff); /* draw figures */ draw((-2,-4)--(8.35,4.76)); draw((10,-4)--(-2,-4)); draw((8.35,4.76)--(10,-4)); draw((9.22,0.11)--(-2,-4)); draw((3.93,1.03)--(10,-4)); draw((-2,-4)--(9.59,-1.81)); draw((10,-4)--(4.99,1.92)); /* dots and labels */ dot((8.35,4.76),dotstyle); label("$C$", (8.48,4.96), NE * labelscalefactor); dot((-2,-4),dotstyle); label("$A$", (-2.63,-4.16), NE * labelscalefactor); dot((10,-4),dotstyle); label("$B$", (10.12,-3.8), NE * labelscalefactor); dot((3.93,1.03),dotstyle); label("$E$", (3.45,1.07), NE * labelscalefactor); dot((9.22,0.11),dotstyle); label("$D$", (9.35,0.33), NE * labelscalefactor); dot((6.32,-0.95),dotstyle); label("$I$", (6.47,-0.74), NE * labelscalefactor); dot((9.59,-1.81),dotstyle); label("$N$", (9.86,-1.68), NE * labelscalefactor); dot((4.99,1.92),dotstyle); label("$M$", (4.52,2.31), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Because $\angle MEB = \angle NDA$, $C$, $E$, $I$, and $D$ are concyclic. Let $\angle IAB = \alpha$ and $\angle IBA = \beta$. Then $\angle EIA = \angle DIB = \alpha + \beta$, so $\angle CEI = \angle EAI + \angle EIA = 2\alpha + \beta$, and similarly $\angle CDI = \alpha + 2\beta$. Because $CEID$ is cyclic, $\angle CEI + \angle CDI = 180^\circ$, so $3\alpha + 3\beta = 180^\circ$, or $\alpha + \beta = 60^\circ$. Then $\angle C = 180^\circ - 2\alpha - 2\beta = \boxed{60^\circ}$.

From the Law of sines in triangles $ADC$ and $BEC$ we learn that $\sin \angle ADC = \sin \angle BEC$. As the triangle is not isosceles, the angles are not equal and so $CEDI$ is cyclic. Then $\gamma + \left(90^\circ + \frac12 \gamma \right) = 180^\circ$ and $\gamma = 60^\circ$.