Quadrilateral from adjacent sides

Adjacent sides share a vertex. In quadrilateral $ABCD$, sides $AB$ and $BC$ are adjacent because they meet at $B$. Given their lengths and the included angle $\angle B$ between them, points $A$, $B$, and $C$ can be plotted right away: draw $AB$, then at $B$ turn through $\angle B$ and mark off $BC$ along the new ray. Three of the four vertices exist before a single length for $AD$ or $CD$ is ever used.

The other two angles aim rays, not lengths. To close the figure, the problem supplies $\angle A$ (between $AB$ and $AD$) and $\angle C$ (between $CB$ and $CD$). Each of these angles is constructed at a vertex that is already fixed, and each produces a ray pointing toward the missing vertex $D$ — but with no distance marked on it, because $AD$ and $CD$ were never given. The two rays are drawn generously long and left to meet on their own.

Vertex $D$ is wherever the two rays cross. This is the one genuinely new idea in this construction method:

where: $\angle B$ is the angle included between the two given sides, $\angle A$ and $\angle C$ are the two remaining given angles, and $D$ is the intersection point of the rays they generate. Contrast this with the "3 sides + 2 angles" method (built earlier in the same chapter), where every vertex is found by measuring a known length along a ray with a compass. Here the last vertex carries no pre-measured length at all — $AD$ and $CD$ are outputs of the drawing, not inputs to it.

Locating $D$ by intersecting two rays feels like it could produce almost any shape, so it is worth confirming that the result is not arbitrary: once $\angle A$, $\angle B$, and $\angle C$ are chosen, the fourth angle $\angle D$ of the finished quadrilateral is already determined, and checking it against that prediction is the standard way to catch a construction error before submitting the answer.

Step 1. Every simple quadrilateral, convex or not, has interior angles that add to $360^\circ$ — this is the angle-sum property proved earlier in the course by splitting the quadrilateral into two triangles along a diagonal, each contributing $180^\circ$. So for any quadrilateral $ABCD$, $\angle A + \angle B + \angle C + \angle D = 360^\circ$ is guaranteed, regardless of the particular side lengths or which construction method built it.

Step 2. In this construction, $\angle A$, $\angle B$, and $\angle C$ are chosen by the problem and drawn exactly with a protractor before $D$ is even found. The angle-sum property therefore predicts $\angle D$ in advance, purely by subtraction: $\angle D = 360^\circ - (\angle A + \angle B + \angle C)$, with no drawing required to know this value.

Step 3. After the rays from $A$ and $C$ are drawn and their crossing point $D$ is marked, the actual angle at $D$ in the finished figure can be measured with a protractor. If the construction was accurate, that measured angle will match the predicted $\angle D$ from Step 2, typically within a degree or two of drawing error.

Step 4. This gives a genuine, independent check: the predicted $\angle D$ comes from arithmetic done before any ray is drawn, while the measured $\angle D$ comes from the physical intersection point. Agreement between the two confirms that the rays were opened on the correct side of each side line and extended far enough to meet correctly — a mismatch of more than a couple of degrees means a ray was likely drawn in the wrong direction.

Construct quadrilateral $ABCD$ in which $AB = 5.0$ cm, $BC = 4.5$ cm, $\angle A = 100^\circ$, $\angle B = 80^\circ$, and $\angle C = 95^\circ$. Predict $\angle D$ before constructing, then verify it.

Step 1. Predict $\angle D$ using the angle-sum property before drawing anything: $\angle D = 360^\circ - (\angle A + \angle B + \angle C) = 360^\circ - (100^\circ + 80^\circ + 95^\circ) = 360^\circ - 275^\circ = 85^\circ$.

Step 2. Draw side $AB = 5.0$ cm with a ruler, then at $B$ construct $\angle ABX = 80^\circ$ with a protractor, where $BX$ is the ray for side $BC$. Along $BX$, mark $BC = 4.5$ cm to fix point $C$. Points $A$, $B$, $C$ are now plotted.

Step 3. At $A$, construct $\angle BAY = 100^\circ$, where $AY$ is a ray toward $D$; draw it long. At $C$, construct $\angle BCZ = 95^\circ$, where $CZ$ is a ray toward $D$; draw it long too. Mark the point where $AY$ and $CZ$ cross as $D$, and join $AD$ and $CD$.

Step 4. Measure the constructed $\angle D$ with a protractor. It should read close to the Step 1 prediction of $85^\circ$, confirming the construction is accurate. Final answer: $ABCD$ is completed with $\angle D \approx 85^\circ$, matching the angle-sum prediction, and all four angles sum to $360^\circ$.

Competitive stretch

If instead $AB = 4.0$ cm, $BC = 5.5$ cm, $\angle P{=}A = 85^\circ$, $\angle B = 90^\circ$, and $\angle C = 100^\circ$ (the second Pass-1 example, relabelled), the same subtraction gives $\angle D = 360^\circ - (85^\circ + 90^\circ + 100^\circ) = 85^\circ$ before drawing a single ray — a useful habit before every construction of this type, since it turns the final measurement into a check rather than a guess.

• Drawing the two "free" rays too short, so they never meet on the page — always extend rays generously since $D$ is found only by intersection.
• Opening the protractor on the wrong side of a side line, which sends a ray away from where it should meet the other ray and produces no valid quadrilateral at all.
• Marking off a length for $AD$ or $CD$ with a compass — these lengths are not given; they must emerge naturally from where the rays cross.
• Skipping the angle-sum prediction, which is the fastest way to catch a construction error: compute $\angle D = 360^\circ - (\angle A+\angle B+\angle C)$ first, then confirm the measured angle matches.

Ask the tutor: Want a side-by-side comparison of all four quadrilateral-construction methods in this chapter, so you can spot which one to use from the given data alone?