Quadrilateral from sides and diagonals

A compass cannot draw a four-sided figure directly — it only draws arcs, and arcs only fix points at a known distance from a known centre. So every quadrilateral construction is secretly a triangle construction done twice. When you are given three sides and both diagonals of quadrilateral $ABCD$ , the trick is to notice that three of those five given lengths — two sides and the diagonal joining their outer ends — already describe a complete triangle by SSS. Build that triangle first, and three of the four vertices are fixed. The fourth vertex is then trapped between two circles: one centred at a known vertex with the remaining side as radius, one centred at another known vertex with the remaining diagonal as radius. Where those two arcs cross is the last corner. This article expands the Class 8 "3 sides + 2 diagonals" construction from the Pass-1 note into a full canon-lite treatment: the general rule stated algebraically, a worked construction carried through with real measurements, and the traps that turn a clean quadrilateral into a crossed or impossible one.

The core insight

Three given sides and one diagonal build a first triangle by SSS, and the fourth vertex is then located as the intersection of two arcs drawn from the remaining side and the remaining diagonal.

The core idea

Every construction problem in this chapter reduces to the one tool a compass and straightedge give you for free: SSS triangle construction, where three known lengths fix a triangle uniquely (up to a mirror flip). A quadrilateral has four vertices, so it needs one triangle construction to fix the first three and a second "two arcs meet at a point" step to fix the fourth.

Reading the given data. In quadrilateral $ABCD$ you are given three sides — commonly $AB$ , $BC$ , $CD$ — plus both diagonals $AC$ and $BD$ . Five lengths in total. Look for the group of three that already forms a triangle: here $AB$ , $BC$ , and diagonal $AC$ describe triangle $ABC$ completely, since $AC$ is the segment joining $A$ and $C$ across the quadrilateral.

Fixing the first three vertices. Constructing $\triangle ABC$ by SSS gives you $A$ , $B$ , and $C$ as actual points on the paper, in their correct relative positions. Nothing about vertex $D$ has been used yet.

Trapping the last vertex. Vertex $D$ has to be, at the same time, at distance $CD$ from $C$ (the leftover side) and at distance $BD$ from $B$ (the leftover diagonal). Each of those conditions alone describes a whole circle of possible positions; together, two circles meet in at most two points, and the correct one is chosen by staying on the correct side of the figure (opposite $B$ , so the quadrilateral does not fold back on itself). This is exactly the same "distance from two fixed centres" logic used to build a triangle from three sides — applied a second time, with $C$ and $B$ as the two centres instead of two triangle vertices.

D = \operatorname{arc}(C,\, CD) \cap \operatorname{arc}(B,\, BD)

where: $\operatorname{arc}(C, CD)$ is the set of points at distance $CD$ from centre $C$ , and $\operatorname{arc}(B, BD)$ is the set of points at distance $BD$ from centre $B$ ; their intersection on the far side of $AC$ from $B$ is the vertex $D$ .

Figure

Quadrilateral ABCD built from triangle ABC by SSS then vertex D found where an arc from C of radius CD meets an arc from B of radius BD

Why the two-arc idea always works

Derivation ALocating the fourth vertex of a quadrilateral from three sides and two diagonals

The goal is to show, using only the SSS triangle-construction principle a student already has from Class 7, why fixing triangle $ABC$ first and then intersecting two arcs at $C$ and $B$ correctly and uniquely places the fourth vertex $D$ , rather than treating the last step as a rule to memorise without justification.

Step 1. Identify which three of the five given lengths ( $AB$ , $BC$ , $CD$ , $AC$ , $BD$ ) name a genuine triangle. A triangle needs three segments that close up into one figure — here $A$ , $B$ , $C$ are three points, and $AB$ , $BC$ , $AC$ are exactly the three sides joining them pairwise, so this triple is constructible by SSS the moment each pair of lengths satisfies the triangle inequality.

Step 2. Construct $\triangle ABC$ : draw $AB$ as a base segment, swing an arc of radius $AC$ from $A$ and an arc of radius $BC$ from $B$ , and mark their intersection as $C$ . This is standard SSS construction, and it fixes $A$ , $B$ , $C$ as actual coordinates on the page — three of the quadrilateral's four vertices are now known exactly.

Step 3. Notice that the two unused lengths, $CD$ and $BD$ , are both distances measured from already-fixed points ( $C$ and $B$ ) to the unfixed point $D$ . A single distance constraint from a fixed point describes a whole circle of candidate positions — not yet a unique point. But two independent distance constraints, from two different fixed centres, intersect in at most two points, because two distinct circles meet in zero, one, or two points.

Step 4. Draw both circles as arcs: one centred at $C$ with radius $CD$ , one centred at $B$ with radius $BD$ . Take the intersection point on the side of line $AC$ opposite $B$ — this side condition is what keeps $ABCD$ a simple (non-crossed) quadrilateral rather than folding vertex $D$ back into the triangle's interior. That intersection point is $D$ , and by construction it is simultaneously at distance $CD$ from $C$ and distance $BD$ from $B$ , exactly matching the two given lengths.

Step 5. Since every step used only lengths that were part of the original given data, and the two arcs were drawn from points already fixed by SSS, the resulting figure $ABCD$ has all three given sides, both given diagonals, and no length was assumed or estimated. This is why the "triangle first, two arcs second" procedure is not a special trick for this one problem but the general method for any "3 sides + 2 diagonals" construction.

Worked example

Construct quadrilateral $PQRS$ with $PQ = 3.6$ cm, $QR = 4.4$ cm, $RS = 3.9$ cm, diagonal $PR = 5.8$ cm, and diagonal $QS = 5.5$ cm.

Step 1. Spot the constructible triangle. The three given lengths $PQ$ , $QR$ , and diagonal $PR$ name triangle $PQR$ completely, so this is the triangle to build first. Sketch the quadrilateral roughly to confirm $S$ will sit on the far side of $PR$ from $Q$ .

Step 2. Construct $\triangle PQR$ by SSS. Draw $PQ = 3.6$ cm with a ruler. With centre $P$ and radius $5.8$ cm, draw an arc for diagonal $PR$ . With centre $Q$ and radius $4.4$ cm, draw an arc for side $QR$ , cutting the first arc at $R$ . Check the triangle inequality: $3.6 + 4.4 = 8.0 > 5.8$ , so the triangle is valid. Join $PR$ and $QR$ .

Step 3. Locate $S$ with two more arcs. With centre $R$ and radius $3.9$ cm (side $RS$ ), draw an arc on the side of $PR$ opposite $Q$ . With centre $Q$ and radius $5.5$ cm (diagonal $QS$ ), draw a second arc cutting the first at $S$ .

Step 4. Join $RS$ , $SP$ , and the diagonal $QS$ as a dashed construction line to display both diagonals. The completed quadrilateral $PQRS$ has $PQ = 3.6$ cm, $QR = 4.4$ cm, $RS = 3.9$ cm, $PR = 5.8$ cm and $QS = 5.5$ cm, with vertex $S$ fixed uniquely by the two arcs from $R$ and $Q$ .

Competitive stretch

If a fifth given length were the side $SP$ instead of one of the diagonals, the same "two arcs" idea still applies — you would just be intersecting an arc from $R$ (radius $RS$ ) with an arc from $P$ (radius $SP$ ) instead of from $R$ and $Q$ . The identity of the two centres always comes from whichever two vertices are already fixed and have a known distance to the missing vertex; only the labels change.

Common traps

Picking the wrong starting triangle. Only the triple of lengths that names two adjacent given sides plus the diagonal joining their outer vertices is directly constructible first. A rough sketch before picking up the compass avoids this.
Swapping the centres for the last two arcs. The fourth vertex needs one arc from the leftover side's far endpoint and one arc from the leftover diagonal's far endpoint — mixing up which length goes with which centre gives a wrong point.
Drawing the second pair of arcs on the wrong side of the first triangle, which produces a self-intersecting (crossed) quadrilateral instead of a simple one.
Skipping the triangle inequality check, both for the first triangle built directly and, implicitly, for the arcs at the fourth vertex — if the two radii and the already-drawn diagonal cannot form a triangle, the arcs will not meet at all.

Ask the tutor: Want to see how this construction changes when one of the diagonals is replaced by an angle instead, as in the "3 sides + 2 angles" method?

What if none of the three given sides plus a diagonal happens to form a valid triangle

Then the data is inconsistent and no such quadrilateral exists. Always check the triangle inequality on your chosen triple, such as $3.6 + 4.4 = 8.0 > 5.8$ for triangle $PQR$ above, before drawing a single arc. If the sum of the two shorter lengths does not exceed the third, the arcs for that triangle will never meet, and you should re-examine whether the given measurements were copied correctly.

Does it matter which diagonal I treat as belonging to the first triangle

Yes. Only the diagonal that joins the two endpoints of your two chosen adjacent sides can complete that first triangle — for triangle $PQR$ built from sides $PQ$ and $QR$ , the closing segment is diagonal $PR$ , not $QS$ . The other diagonal, $QS$ , is saved for the second stage, where it fixes vertex $S$ together with the remaining side $RS$ .

How is this different from constructing a quadrilateral from four sides and one diagonal

In "4 sides + 1 diagonal" the single given diagonal splits the quadrilateral into two triangles that share that diagonal as a common side, and each triangle is built by SSS in turn using two sides plus the shared diagonal. In "3 sides + 2 diagonals," only one triangle is directly built from given data (two sides plus one diagonal); the fourth vertex is not built from a second full SSS triangle but trapped by two arcs, one from a side length and one from the second diagonal, meeting at a point.