The equation gives $\frac{bc}{a+c}+\frac{ac}{b+c}=c$. Simplyfing this equation, we get $c^2=a^2+b^2-ab$, hence $\angle C=60^{\circ}$.

Did you get the last part from Law of Cosines: $c^{2}=a^{2}+b^{2}-2ab\cos{\theta}$ and the fact that 2$\cos{\theta}=1$

Yes. It's quite well-known, so I skipped the proof.

I figured you did...

Another method. Let $F$ be on $AB$ such that $AE=AF$, then $ED=EF$. Let incenter be $I$. Then $\angle IEA=\angle IFA=\angle IDC$ so $CDIE$ is cyclic, so $\angle ACB=180-\angle AIB$. But $\angle AIB=90+\frac12 \angle ACB$. Solving gives $\angle C=60$.

WakeUp wrote: The bisectors of the angles $A$ and $B$ of the triangle $ABC$ meet the sides $BC$ and $CA$ at the points $D$ and $E$, respectively. Assuming that $AE+BD=AB$, determine the angle $C$. by angle bisector theorem we deduce $ \frac{b}{c+a}+\frac{a}{c+b}=1$ and by expnding we have $a^2+b^2-ab=c^2$ no by cosine law $\cos\gamma=\frac{1}{2}$ so $\gamma=60^{\circ}$

Here's the geometric solution Let E1 and E2 be the reflection of E by AD and BE Because AE+BD=AB, E1 and E2 be the same point. So $\angle AIB$=120 and $\angle C$=60

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