Given the length of each side, construct a quadrilateral $ABCD$ such that diagonal $AC$ bisects $\angle A$.
(Source: Kiselev's Geometry, Vol 1, Ex 220.)
My solution is below. Please verify, critique, improve, or help me with the gap (marked *).
Construction: Assume WLOG that $AD \geq AB$. Draw $AB$ and extend it to $D'$ such that $AD' \cong AD$. Draw a circle with center $B$ and radius $BC$, and a circle with center $D'$ and radius $CD$. Label a point of intersection of these circles as $C$.
Reflect $D'$ over line $AC$ and label it $D$. Then $ABCD$ is the desired quadrilateral.
Proof: Recall that for any point $P$, line $\ell$, and point $L$ on $\ell$, if $P'$ is the reflection of $P$ over $\ell$, then $PL \cong P'L$ by SAS. Therefore, since $A$ and $C$ are on line $AC$, $AD \cong AD'$ and $CD \cong CD'$. Thus each side is of the correct length, and $\angle DAB \cong \angle DAD'$ is bisected by $AC$ by SSS.
Boundaries: If $BC + CD > BD' = AD - AB$, the circles will intersect at a point not on the line $AB$, as assumed. If $BC + CD = AD - AB$, the intersection will be on the line, and the quadrilateral will be degenerate but still correct. And if $BC + CD < AD - AB$, no such quadrilateral exists ([*] How do I show this?)
Discussion: The core challenge is that while we can immediately construct any two sequential points of the quadrilateral, the third and fourth are then only specified by non-intersecting circles. We solve this as soon as we realize that since $AC$ bisects $\angle BAD$, then $A,B,D'$ are colinear. Moreover, we don't need to wait until we've drawn $AC$ to draw $D'$ - we can draw a single line $ABD'$ with three points from the start. The rest of the construction immediately follows.