Construction of a triangle, given: side, sum of the other sides and angle between them.

Given: $\overline{AB}$, $\overline{AC}+\overline{BC}$ and $\angle C$. Construct the triangle $\triangle ABC$ using rule and compass.

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Let fix the points $A$, $B$, then $C$ is an intersection point of the arc of circle where the segment $\overline{AB}$ is seen with the given angle and the ellipse with foci $A$, $B$ and distance $\overline{AC}+\overline{BC}$. –  Sgernesto Feb 7 '13 at 21:47

Let's see... Given a segment $\overline{AB}$ and an angle $\alpha$, the locus of points $P$ such that $\angle APC=\alpha$ is an arc of a circle, which you can draw, since you can find at least one such point $P$ (namely, the point for which the lengths of $\overline{AP}$ and $\overline{PB}$ coincide; you can draw this point $P$ as $\angle PAB=\angle PBA=(\pi-\alpha)/2$).
You know $\angle C$, so draw the corresponding arc of a circle $\mathcal C_1$. Also, draw the arc of a circle $\mathcal C_2$ consisting of all points $P$ with $\angle APB=\angle C/2$. Also, draw the circle $\mathcal C_3$ with center $A$ and radius the length $\overline {AC}+\overline{CB}$.
This circle $\mathcal C_3$ meets $\mathcal C_2$ at a point $P$. The segment $\overline{PA}$ meets the circle ${\mathcal C}_1$ at $C$.
To see this, call $D$ the point of intersection of $\overline{PA}$ and ${\mathcal C}_1$. Then $\angle ADB=2\angle APB$, so $\angle DBP=\angle APB=\angle DPB$, so $\triangle BDP$ is isosceles, and the lengths of $\overline{DP}$ and $\overline{DB}$ coincide, so the length $\overline{AD}+\overline{DB}$ is the same as the length of $\overline{AD}+\overline{DP}$, that is, the length of $\overline{AP}$, which is the given length $\overline{AC}+\overline{CB}$. Also, $\angle ADB=\angle C$, by construction.