Constructing a parallelogram, given its side, the sum of the diagonals, and the angle between them

Question

Kiselev's Planimetry book, exercise 216:

Construct a parallelogram, given its side, the sum of the diagonals, and the angle between them.

Can one provide a hint for this?

An equivalent formulation is:

Construct a triangle, given its base, the angle of the vertex, and the sum of the two other sides.

My initial idea was to construct some triangle given the angle and the two given lengths, but this did not get me anywhere. Also, there are some solutions for this book on wikidot.com, but this one is not included.

Answer 1

As you imply, I'll tackle this with the triangle formulation, as it is a little more natural.

Let $\overline{AP}$ be the sum of the two sides, and $\alpha$ the measure of the vertex angle. Consider some arbitrary point $B$ on $\overline{AP}$, which will be the apex of our triangle. Then construct $C$ such that $m\angle ABC=\alpha$ and $BC=BP$. So our goal is to choose $B$ in such a way that $AC$ is the given base length.

What is the locus of all possible points $C$ as $B$ varies along $\overline{AP}$? (Hint: what we can we say about $m\angle APC$?)

As $\angle APC$ is an inscribed angle that intercepts the same arc as central $\angle ABC$, $m\angle APC=\frac\alpha 2$. As that angle is independent of the location of $B$, the locus of possible points $C$ is a ray extending from $P$ such that $\alpha APC=\frac\alpha 2$.

Given that, constructing the triangle is straightforward.

Construct the aforementioned ray, then construct a circle with center $A$ and radius the given base length. Take $C$ to be the first point where the ray meets the circle. Then construct the perpendicular bisector of $\overline{PC}$ and take $B$ to be the intersection of that line with $\overline{AP}$. Then $\triangle ABC$ is the desired triangle.