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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 $\angle APB=\alpha$ 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.

(You probably want to embellish this by counting how many solutions are obtained this way, and checking that all solutions come from this construction, but I expect this should be more or less direct now.)

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Draw the side given AB draw the angle given A produce the adjacent side which equal to sum of given sides AP connect remaining point from it PB bisect that side PB produce that side until cut AP take the point of intersection C now you have triangle. enter image description here

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Let $a = BC$, $b = CA$, $c = AB$, $\gamma = \angle C$. We are given $\gamma$, $c$, $a + b$.

Construct an angle of measure $\gamma$ at a vertex $C$. The problem is to place $A$ and $B$ on the sides of the angle.

Construct the semi-perimeter $s = \frac{1}{2}[(a+b) + c]$. The points of contact $D_2$ and $E_2$ of the excircle of $ABC$ relative to vertex $C$ with sides $CA$ and $CB$ are at a distance $CD_2 = CE_2 = s$ from $C$. Place these points on the side of the angle and construct the excircle.

Similarly, the points of contact $D_1$ and $E_1$ of the incircle of $ABC$ with sides $CA$ and $CB$ are at a distance $CD_1 = CE_1 = s-c$ from $C$. Place these points on the side of the angle and construct the incircle.

Now the problem is reduced to that of constructing a common internal tangent to two circles, whose solution is well known.

enter image description here

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