Constructing a triangle given two angles and the sum of two sides

Question

I am looking for an answer to the following construction

Construct a triangle given two angles (3 angles) and the sum of two sides

Answer 1

Step 1: Draw a line $AB$ of arbitrary length. Copy angle $\alpha$ at point $A$ and angle $\beta$ at point $B$ to form $\triangle ABC$.

Step 2: Extend line $AB$ in the direction of $B$ by length $AC$, calling that line $AD$.

Step 3: Draw line a perpendicular to $AD$ at $A$, and mark off the given required sum of two sides $s$ such that the length of $AE$ is $s$. Connect points $D$ and $E$ to form line $DE$.

Step 4: construct a line parallel to $DE$ passing through point $B$. This line intersects line $AE$ at some point $F$.

Step 5: Copy angle $\alpha$ such that one leg is on $AF$ and the vertex is at $A$. Copy angle $\alpha$ such that one leg is on $AF$ and the vertex is at $A$. Copy angle $\beta$ such that one leg is on $AF$ and the vertex is at $F$. The two new lines meet at some point $G$.

Triangle $AFG$ is the required triangle.

Note that the solution is not unique, unless the problem has specified that the sum of two sides meeting at a specific one of the angles must be the given sum.

Answer 2

Let us suppose the problem is solved and $\triangle ABC$ is our desired triangle. Extend $AB$ from $B$ to point $H$ such that $|BH| = |BC|$ and so $|AH|=|AB|+|BC|$. Note that $\widehat H = \frac12 \widehat{ABC}$ (why?)

Now let us suppose the problem is not solved. We only have the three angles of $\triangle ABC$ and the sum of lengths $|AB|+|BC|$. We can construct $\triangle AHC$ (why?) and you can surely do the rest.