Quadrilaterals defined

A quadrilateral is a closed plane figure bounded by four line segments (sides), meeting at four points called vertices, and containing four interior angles. If the vertices, taken in order around the boundary, are $A$, $B$, $C$, $D$, the quadrilateral is written $ABCD$, with sides $AB$, $BC$, $CD$, $DA$.

The parts, precisely. Four labels do most of the work in this chapter, and every later theorem leans on them being unambiguous:

• Sides are the four boundary segments — $AB, BC, CD, DA$ — and vertices are the four corners where consecutive sides meet.
• Adjacent sides share a vertex (e.g. $AB$ and $BC$ meet at $B$); opposite sides do not (e.g. $AB$ and $CD$).
• Adjacent angles sit at vertices joined by a side (e.g. $\angle A$ and $\angle B$); opposite angles sit at vertices with no shared side (e.g. $\angle A$ and $\angle C$).
• Diagonals join opposite, non-adjacent vertices — $AC$ and $BD$. Every simple quadrilateral has exactly two.

Simple, closed, and convex or concave. For a four-sided figure to be a quadrilateral in this chapter's sense, it must be simple — its sides do not cross each other before the boundary closes — and it must genuinely close, returning to the starting vertex. A simple quadrilateral is further either convex, when both diagonals lie fully inside the figure, or concave, when one diagonal strays partly outside because one interior angle is reflex (greater than $180°$). A figure whose sides cross before closing is self-intersecting, and it is not a quadrilateral of this kind at all — it is excluded from the chapter entirely.

Vertex order fixes the naming. Quadrilateral $PQRS$ names its vertices in order around the boundary, clockwise or anticlockwise, so the sides are exactly $PQ$, $QR$, $RS$, $SP$. Writing the same four points as $PRQS$ describes a different figure — usually a self-intersecting one — because $PR$ and $QS$ would then be read as sides, swapping roles with what were the diagonals. This single convention is why every worked problem in the chapter insists on reading vertex letters strictly in the order given.

The angle-sum fact needs no new tools beyond what one diagonal reveals, so this section reasons it through directly rather than deriving a formula from scratch. The claim is precise enough to state as a target before proving it: for every simple quadrilateral $ABCD$, however irregular its four sides look, the four interior angles $\angle A + \angle B + \angle C + \angle D$ always add up to exactly $360°$, and the reason is that any diagonal splits the quadrilateral into exactly two triangles whose angle sums are each independently fixed at $180°$ by the triangle angle-sum property already known from earlier classes.

Step 1. Draw the diagonal $AC$ in quadrilateral $ABCD$. Because $ABCD$ is simple, this diagonal lies inside the figure (for a convex quadrilateral) or at least connects two non-adjacent vertices cleanly, and it splits the four-sided region into exactly two triangles: $\triangle ABC$ and $\triangle ACD$. Nothing about the sides or angles has changed — the diagonal has simply been added as a new segment, so the quadrilateral's boundary and the two triangles together cover the same region with no overlap and no gap.

Step 2. Apply the triangle angle-sum property to each piece separately. In $\triangle ABC$, the three angles sum to $180°$: $\angle BAC + \angle B + \angle BCA = 180°$. In $\triangle ACD$, the three angles likewise sum to $180°$: $\angle DAC + \angle D + \angle DCA = 180°$.

Step 3. Add the two equations and regroup by vertex. $(\angle BAC + \angle DAC) + \angle B + (\angle BCA + \angle DCA) + \angle D = 360°$. But $\angle BAC$ and $\angle DAC$ together make up the full angle $\angle A$ of the quadrilateral (the diagonal splits $\angle A$ into exactly those two pieces), and $\angle BCA$ and $\angle DCA$ together make up the full angle $\angle C$. Substituting gives $\angle A + \angle B + \angle C + \angle D = 360°$.

where: $\angle A, \angle B, \angle C, \angle D$ are the four interior angles of quadrilateral $ABCD$, taken at the vertices in order. This result holds for every simple quadrilateral — convex or concave — because any diagonal, drawn from any vertex, splits it into exactly two triangles whose two independent $180°$ sums always combine to $360°$; the same argument works starting from diagonal $BD$ instead of $AC$.

Problem. In quadrilateral $MNOP$, list (a) all four sides, (b) both diagonals, (c) the pair of angles opposite to $\angle M$, (d) two pairs of adjacent sides, and (e) given $\angle M = 100°$, $\angle N = 80°$, $\angle O = 95°$, find $\angle P$.

Step 1. Since the vertices are named in order $M, N, O, P$, the sides are formed by consecutive letters, wrapping back to the start: $MN$, $NO$, $OP$, $PM$. That answers (a).

Step 2. Diagonals join non-adjacent (opposite) vertices. In $MNOP$, $M$ and $O$ are opposite, and $N$ and $P$ are opposite, so the diagonals are $MO$ and $NP$. That answers (b).

Step 3. $M$'s adjacent vertices are $N$ and $P$ (each shares a side with $M$), so the vertex opposite $M$ is $O$, and the angle opposite $\angle M$ is $\angle O$. That answers (c). For (d), two valid adjacent-side pairs are $MN$ and $NO$ (sharing vertex $N$), and $OP$ and $PM$ (sharing vertex $P$).

Step 4. Using the angle-sum result from the reasoning above, $\angle M + \angle N + \angle O + \angle P = 360°$. Substituting the given values, $100° + 80° + 95° + \angle P = 360°$, so $275° + \angle P = 360°$, giving $\angle P = 85°$.

Answer: sides $MN, NO, OP, PM$; diagonals $MO$ and $NP$; angle opposite $\angle M$ is $\angle O$; adjacent pairs $(MN, NO)$ and $(OP, PM)$; and $\angle P = 85°$.

Competitive stretch

Problem. A quadrilateral $ABCD$ has $\angle A = 100°$, and when diagonal $AC$ is drawn it lies completely inside the figure, splitting it into $\triangle ABC$ and $\triangle ACD$. Is $ABCD$ convex or concave, and how many diagonals does it have in total?

Step 1. Recall the definitions: a quadrilateral is convex if both diagonals lie entirely inside it, and concave if at least one diagonal passes outside because some interior angle is reflex.

Step 2. Diagonal $AC$ is given to lie fully inside, $\angle A = 100°$ is not reflex, and nothing in the problem suggests any other angle is reflex or that diagonal $BD$ would exit the figure. That evidence is consistent with $ABCD$ being convex.

Step 3. Every simple quadrilateral, convex or concave, has exactly 2 diagonals — here $AC$ and $BD$ — since a diagonal joins each pair of non-adjacent vertices, and a four-vertex figure has exactly two such pairs.

Answer: $ABCD$ is convex, with 2 diagonals, $AC$ and $BD$.

• Confusing "adjacent" with "opposite." Always check whether the two sides or two angles share a vertex (adjacent) or not (opposite) before answering.
• Treating a diagonal as a side. A diagonal joins non-adjacent vertices and lies inside the figure; it is never one of the four boundary sides.
• Ignoring vertex order when naming. Writing vertices out of order (e.g. $ABDC$ instead of $ABCD$) swaps which segments count as sides versus diagonals, and can describe a self-intersecting figure.
• Forgetting the $360°$ angle sum works for any diagonal. Students sometimes think only one particular diagonal gives the correct split; either diagonal, $AC$ or $BD$, produces two triangles whose sums add to the same $360°$.

Ask the tutor: Want to practice labelling sides, angles and diagonals on a few different quadrilaterals, or work through more angle-sum problems until the vertex-order logic feels automatic? Ask your AI tutor to sketch examples and quiz you.