Class 9 › Mathematics › Chapter

Quadrilaterals

The precise definition of a quadrilateral, convex and non-convex 4-gons, properties and tests for parallelograms, the Midpoint Theorem and its converse, the Centroid Theorem, Varignon's theorem, and tiling the plane with any 4-gon.

1. Tiling and the Hinged Grid

Tiling the plane: covering the plane with copies of a given shape (or shapes) with no gaps and no overlaps.

Holding the bottom stick fixed and pushing the top turns every square tile into a parallelogram tile; the grid still covers the plane.

Easy and hard cases: tiling with a rectangle is easy; with an irregular quadrilateral it seems difficult. The chapter ends by showing that every quadrilateral tiles the plane.
Hinged grid: imagine long sticks forming a rectangular grid, pinned at every crossing so that they can rotate. Hold the bottom stick fixed and push the vertical side.
What changes and what stays: each set of parallel lines stays parallel and the lengths stay the same, but the angle between the two sets is no longer $90^\circ$. Every tile is now a parallelogram.
Conclusion: the plane can be tiled with copies of any fixed parallelogram.
Plan of the chapter: define a quadrilateral precisely, study parallelograms (properties and tests to recognise them), use them to prove facts about triangles, and finally answer the tiling question for every 4-gon.

2. What Exactly is a Quadrilateral?

A triangle first: three points $A, B, C$ give a triangle (the points on segments $AB, BC, CA$) only if the points are non-collinear. If they are collinear we only get a line segment. Non-collinear $A, B, C$ are the vertices and the three segments are the sides.
For four points: just taking four points and joining $AB, BC, CD, DA$ does not always give the figure we picture. Some care is needed.

Four conditions for a quadrilateral

1. No three vertices collinear. Otherwise two adjacent sides lie on one line, giving a triangle (as in NOPE) or overlapping sides (as in SILY).

2. All four vertices in one plane. Four-sided figures that are not flat (as in BENT) are non-planar quadrilaterals.

3. The sides do not cross. Every point of the figure other than a vertex must lie on exactly one side. A figure whose sides cross (as in CUTS) is called a self-intersecting quadrilateral and is not counted as an ordinary quadrilateral.

4. The figure is closed. All four segments $AB, BC, CD, DA$ must be present, so the sides join end to end in a loop. An open figure (as in OPENS) is not a quadrilateral.

Definition of a quadrilateral: let $A, B, C, D$ be four distinct points in a plane. The points on the four segments $AB, BC, CD, DA$ form the quadrilateral $ABCD$ if every such point other than $A, B, C, D$ lies on exactly one of these four segments.

Figure	What goes wrong	Quadrilateral?
NOPE	Three vertices are collinear, so it is really a triangle	No
SILY	Three vertices are collinear, so two sides overlap	No
BENT	The four vertices are not in one plane	No (non-planar)
CUTS	The sides cross each other	No (self-intersecting)
OPENS	The figure is not closed	No
DART	Has a dent, but is planar and does not cross itself	Yes (non-convex)

Convention: unless stated otherwise, "quadrilateral" means a planar, non-self-intersecting quadrilateral. A quadrilateral is also called a 4-gon.

3. Terms and Naming

Term	Meaning (in quadrilateral ABCD)
Vertices	The points $A, B, C, D$
Sides (edges)	The segments $AB, BC, CD, DA$
Diagonals	The segments $AC$ and $BD$, joining vertices that are not adjacent
Adjacent vertices	The two endpoints of a side
Adjacent sides	Two sides with a common endpoint
Opposite sides	Two sides with no common endpoint
Internal angles	Four angles, each formed at a vertex by the pair of adjacent sides there
Adjacent angles	Internal angles at adjacent vertices
Opposite angles	Internal angles at vertices that are not adjacent (the two ends of a diagonal)

Examples of the terms: side $AB$ is adjacent to $BC$ and $DA$, and opposite to $CD$. Angle $\angle A$ is adjacent to $\angle B$ and $\angle D$, and opposite to $\angle C$.
Naming: the vertices must be named in the order in which they are met while travelling along the sides. Quadrilateral $ABCD$ can also be called $BCDA$, $CDAB$, $DABC$, $DCBA$, $ADCB$, $BADC$ or $CBAD$ (start anywhere, go either way round), but not by any other order such as $ABDC$.
Internal angle at a vertex: the angle between the two sides at that vertex, measured through the inside of the figure. In a non-convex quadrilateral, the internal angle at the dent is more than $180^\circ$.

4. Convex and Non-convex Quadrilaterals

Convex quadrilateral: a quadrilateral with no dent, that is, all of its internal angles are less than $180^\circ$. A quadrilateral with an internal angle greater than $180^\circ$ is non-convex.

DART looks like triangle ART with a bite taken out; the angle at the dent is more than $180^\circ$.

Test 1 (angles): convex $\iff$ every internal angle is less than $180^\circ$.
Test 2 (diagonals): the diagonals of a convex quadrilateral intersect; those of a non-convex quadrilateral do not.
Four-sided figures (4-gons): they are of four types: non-planar (skew), self-intersecting planar, planar non-self-intersecting non-convex, and convex. The last two are the "usual" quadrilaterals.
Angle sum: the four internal angles of a planar, non-self-intersecting quadrilateral add up to $360^\circ$, even if it is non-convex. For a self-intersecting or a non-planar quadrilateral the sum is less than $360^\circ$.
Parallel sides force convexity: if $AB \parallel DC$ in a quadrilateral $ABCD$, then it cannot be non-convex, so its diagonals always intersect. This fact is used in Theorem 5.
Why odd 4-gons matter: in a 4-bar linkage (bars joined at their ends, with some ends fixed), the bars can take convex, non-convex and even self-intersecting positions. Engineers must account for all of them.

5. Parallelograms and Theorem 1

Parallelogram: a quadrilateral whose opposite sides are parallel. Rhombuses, rectangles and squares are all parallelograms.

Theorem 1 (properties of a parallelogram) (a) Opposite sides are equal. (b) Opposite angles are equal. (c) The diagonals bisect each other.

Opposite sides and opposite angles are equal; the diagonals cut each other at their midpoint $E$.

Proof of Theorem 1

(a) In parallelogram $ABCD$, the diagonal $AC$ is a transversal for $AB \parallel DC$ and also for $AD \parallel BC$. So $\triangle ACD \cong \triangle CAB$ by the ASA test, which gives $AB = DC$ and $AD = BC$.

(b) $BC \parallel AD$ with $AB$ a transversal, so $\angle A + \angle B = 180^\circ$ (interior angles on the same side). In the same way, any two adjacent angles add to $180^\circ$. Then $\angle A = 180^\circ - \angle B = \angle C$, and similarly $\angle B = \angle D$.

(c) Let the diagonals meet at $E$. Using (a), $\triangle AED \cong \triangle CEB$ (ASA or AAS), so $EA = EC$ and $EB = ED$. Thus $E$ is the midpoint of both diagonals.

Special parallelograms: a parallelogram with one right angle is a rectangle (adjacent angles add to $180^\circ$, so all four angles are $90^\circ$). A parallelogram with equal diagonals is a rectangle.
Diagonal bisecting an angle: if the diagonal $AC$ of a parallelogram $ABCD$ bisects $\angle A$, then $\angle BAC = \angle DAC = \angle ACB$ (alternate angles), so $BA = BC$. Adjacent sides are equal, so $ABCD$ is a rhombus (and $AC$ also bisects $\angle C$).
Perpendicular diagonals: a parallelogram with perpendicular diagonals is a rhombus. It need not be a square, because every rhombus has perpendicular diagonals.
Equal diagonals alone are not enough: a general quadrilateral with equal diagonals need not be a rectangle (an isosceles trapezium is a counter-example). Diagonals that are equal and bisect each other give a rectangle.
Diagonals that bisect each other at right angles make the quadrilateral a rhombus (by Theorem 4 it is a parallelogram, and perpendicular diagonals make it a rhombus).
Angle bisectors: in a parallelogram with unequal adjacent sides, the four angle bisectors meet in pairs at the vertices of a rectangle. In a rhombus they collapse onto the diagonals, which is why the condition of unequal sides is needed.
Ancient Indian view: quadrilaterals were first classified by whether their diagonals are equal or not, before considering equality of sides.

6. Converses of the Parallelogram Properties (Theorems 2 to 4)

Converse: the converse of "If $p$, then $q$" is "If $q$, then $p$". A statement and its converse are separate claims. For example, $x = y \Rightarrow x^2 = y^2$ is true, but the converse is false ($x = 3$, $y = -3$).

Property of a parallelogram	Converse	True?
(a) Opposite sides are equal	If opposite sides of a 4-gon are equal, it is a parallelogram (Theorem 2)	Yes
(b) Opposite angles are equal	If opposite angles of a 4-gon are equal, it is a parallelogram (Theorem 3)	Yes
(c) Diagonals bisect each other	If the diagonals of a 4-gon bisect each other, it is a parallelogram (Theorem 4)	Yes

Strategy: to prove a converse, try running the original proof backwards. This may or may not work, so each backward step must be justified carefully.

Theorem 2: opposite sides equal $\Rightarrow$ parallelogram

In $ABCD$, $AB = DC$ and $AD = BC$. Then $\triangle ACD \cong \triangle CAB$ by SSS, so corresponding angles are equal. Using the transversal $AC$, we get $AB \parallel DC$ and $AD \parallel BC$.

The proof goes through the same congruence $\triangle ACD \cong \triangle CAB$ as the proof of (a), but the test used (SSS instead of ASA) and the conclusion are swapped.

Theorem 3: opposite angles equal $\Rightarrow$ parallelogram

Given $\angle A = \angle C$ and $\angle B = \angle D$: $360^\circ = \angle A + \angle B + \angle C + \angle D = 2(\angle A + \angle B) = 2(\angle B + \angle C)$.

So $\angle A + \angle B = \angle B + \angle C = 180^\circ$. Since $AB$ is a transversal for $BC$ and $AD$, $\angle A + \angle B = 180^\circ$ gives $BC \parallel AD$. Similarly $\angle B + \angle C = 180^\circ$ gives $AB \parallel DC$.

Another form: equal opposite angles make every pair of adjacent angles add to $180^\circ$, and conversely. So Theorem 3 can also be stated: "If each pair of adjacent angles of a quadrilateral is supplementary (adds up to $180^\circ$), then it is a parallelogram."

Theorem 4: diagonals bisect each other $\Rightarrow$ parallelogram

Let the diagonals $AC$ and $BD$ meet at $E$ with $EA = EC$ and $EB = ED$. Also $\angle AED = \angle CEB$ (vertically opposite angles). So $\triangle AED \cong \triangle CEB$ by SAS.

Then $\angle DAE = \angle ECB$, and with the transversal $AC$ this gives $AD \parallel BC$. In the same way $\triangle EAB \cong \triangle ECD$ gives $AB \parallel DC$.

7. The Equal Parallel Sides Test (Theorem 5)

Theorem 5 A quadrilateral with one pair of opposite sides equal and parallel is a parallelogram.

Proof

1. Let $AB \parallel DC$ and $AB = DC$ in quadrilateral $ABCD$. Since $AB \parallel DC$, the quadrilateral is convex, so the diagonals meet at a point $E$.

2. $\angle EAB = \angle ECD$ and $\angle EBA = \angle EDC$ (alternate angles), and $AB = DC$. So $\triangle EAB \cong \triangle ECD$ by ASA.

3. Hence $EA = EC$ and $EB = ED$: the diagonals bisect each other. By Theorem 4, $ABCD$ is a parallelogram.

Test for a parallelogram	Conditions needed
Definition	Both pairs of opposite sides parallel
Theorem 2	Both pairs of opposite sides equal
Theorem 3	Both pairs of opposite angles equal (or all adjacent angle pairs supplementary)
Theorem 4	Diagonals bisect each other
Theorem 5	One pair of opposite sides equal and parallel

Both conditions are needed in Theorem 5: one pair equal alone is not enough (an isosceles trapezium has equal legs), and one pair parallel alone is not enough (a trapezium has one parallel pair).
Useful in practice: if two points $P, Q$ lie on diagonal $BD$ of a parallelogram $ABCD$ with $DP = BQ$, the quadrilateral $APCQ$ is a parallelogram, because its diagonals $AC$ and $PQ$ have the same midpoint.

8. The Midpoint Theorem (Theorem 6)

Theorem 6 (Midpoint Theorem) The segment joining the midpoints of two sides of a triangle is parallel to the third side and has half its length. $$PQ \parallel BC, \qquad PQ = \dfrac{BC}{2}$$

The line $l$ through $C$ parallel to $BA$ meets $PQ$ extended at $R$, producing the parallelogram $BCRP$.

Proof (with an auxiliary construction)

1. In $\triangle ABC$, let $P$ and $Q$ be the midpoints of $AB$ and $AC$. Draw the line $l$ through $C$ parallel to $BA$, meeting line $PQ$ at $R$.

2. In $\triangle APQ$ and $\triangle CRQ$: $AQ = CQ$; $\angle AQP = \angle CQR$ (vertically opposite); $\angle APQ = \angle CRQ$ (since $AP \parallel RC$). So $\triangle APQ \cong \triangle CRQ$ by AAS.

3. Therefore $PQ = QR = \dfrac{PR}{2}$ and $CR = AP$. Since $AP = BP$, we get $CR = BP$.

4. $CR = BP$ and $CR \parallel BP$, so by the equal parallel sides test (Theorem 5), $BCRP$ is a parallelogram. Hence $PR \parallel BC$ and $PR = BC$.

5. Therefore $PQ \parallel BC$ and $PQ = \dfrac{PR}{2} = \dfrac{BC}{2}$.

Break a big problem into one step: proving all three midpoint segments at once is hard because many quantities are unknown. Looking at one segment makes the problem small enough to solve.
Auxiliary construction: drawing the extra line $l$ is a classic geometric trick that comes with experience. A slightly less magical proof extends $PQ$ beyond $Q$ to a point $S$ with $QS = PQ$; then $\triangle APQ \cong \triangle CSQ$ and the same idea finishes the proof.
Cutting a triangle into four: joining the three midpoints $P, Q, R$ of the sides of $\triangle ABC$ splits it into four congruent triangles (by SSS, each has sides equal to half the sides of $\triangle ABC$). Each is a smaller copy of the original triangle.
Rebuilding the triangle: from the middle triangle $PQR$ alone, draw through each vertex a line parallel to the opposite side; these three lines form $\triangle ABC$.

9. Converse of the Midpoint Theorem (Theorem 7)

Theorem 7 The line drawn through the midpoint of one side of a triangle and parallel to another side bisects the third side (and the segment it cuts off is half the parallel side).

Proof 1: running the proof of Theorem 6 backwards

1. $P$ is the midpoint of $AB$, and the line through $P$ parallel to $BC$ meets $AC$ at $Q$. Draw the line through $C$ parallel to $BA$, meeting line $PQ$ at $R$.

2. $BCRP$ has both pairs of opposite sides parallel, so it is a parallelogram and $CR = BP = PA$.

3. $\angle AQP = \angle CQR$ (vertically opposite) and $CR \parallel BA$, so $\triangle APQ \cong \triangle CRQ$ by AAS.

4. Hence $AQ = QC$ (so $Q$ is the midpoint of $AC$) and $PQ = QR$, so $PQ = \dfrac{PR}{2} = \dfrac{BC}{2}$.

Proof 2: using the Midpoint Theorem itself

Let $M$ be the midpoint of $AC$. By Theorem 6, $PM \parallel BC$. Through $P$ there is only one line parallel to $BC$, so the lines $PM$ and $PQ$ coincide. Both meet $AC$ in the same point, so $M = Q$.

Why this is not cheating: using a theorem to prove its converse is allowed here because of one new fact: there is a unique line through a given point parallel to a given line.
Many possible converses: the Midpoint Theorem has two assumptions (P is a midpoint, Q is a midpoint) and two conclusions ($PQ \parallel BC$, $PQ = \frac{BC}{2}$). Different choices give different converses. For example, "$PQ \parallel BC$ and $PQ = \frac{BC}{2}$ $\Rightarrow$ $P$ and $Q$ are midpoints" is true, and Theorem 7 is another. We focus on Theorem 7 because it is particularly useful.
Looking ahead (Grade 10): if only $PQ \parallel BC$ is assumed (without $P$ being a midpoint), the result becomes the Basic Proportionality Theorem: $\dfrac{AP}{PB} = \dfrac{AQ}{QC}$.

10. Medians and the Centroid (Theorem 8)

Median: the segment joining a vertex of a triangle to the midpoint of the opposite side. Every triangle has three medians.

Theorem 8 (Centroid Theorem) The three medians of a triangle pass through one point, which divides each median in the ratio $2 : 1$, the longer part being next to the vertex. This point is the centroid. $$CM : MP = 2 : 1$$

Along the median $CP$: $CM : MP = 2 : 1$. The same ratio holds on the other two medians.

Proof

1. Take two medians $CP$ and $BQ$ meeting at $M$. We show $CM = 2MP$ and $BM = 2MQ$. A theorem for a $2:1$ division is not available yet, so we bisect $CM$ and $BM$ at $Y$ and $X$ and show the three parts of $CP$ are equal.

2. By the Midpoint Theorem in $\triangle ABC$: $PQ \parallel BC$ and $PQ = \dfrac{BC}{2}$. In $\triangle MBC$: $XY \parallel BC$ and $XY = \dfrac{BC}{2}$. So $PQ \parallel XY$ and $PQ = XY$.

3. Then $\triangle MPQ \cong \triangle MYX$ by ASA, so $MQ = MX = XB$ and $MP = MY = YC$. Hence $CM : MP = 2 : 1 = BM : MQ$.

4. Repeat with medians $AR$ and $BQ$ meeting at $N$: $BN : NQ = 2 : 1$. Both $M$ and $N$ divide $BQ$ in the ratio $2 : 1$, and only one point does that, so $M = N$. Thus all three medians pass through $M$.

Concurrent lines: three lines through one common point are called concurrent. Medians are always concurrent.
Direction of the ratio: the longer part ($2$) joins the vertex to the centroid; the shorter part ($1$) joins the centroid to the midpoint of the opposite side.
A second proof (construction): let $M$ be the intersection of medians $CP$ and $BQ$. Extend $AM$ to a point $S$ chosen so that $BSCM$ is a parallelogram. Its diagonals $BC$ and $MS$ bisect each other, so line $AM$ meets $BC$ at the midpoint of $BC$.

11. The Midpoint Theorem for Quadrilaterals (Varignon's Theorem)

Theorem 9 The midpoints of the four sides of a quadrilateral are the vertices of a parallelogram, called the Varignon parallelogram of the original quadrilateral.

$PQRS$ is a parallelogram. Its sides are parallel to the diagonals $AC$ and $BD$ (dashed).

Proof

1. Let $P, Q, R, S$ be the midpoints of $AB, BC, CD, DA$.

2. Midpoint Theorem in $\triangle ABC$: $PQ \parallel AC$. In $\triangle ADC$: $SR \parallel AC$. So $PQ \parallel SR$.

3. Midpoint Theorem in $\triangle BCD$: $QR \parallel BD$. In $\triangle BAD$: $PS \parallel BD$. So $QR \parallel PS$.

4. Both pairs of opposite sides of $PQRS$ are parallel, so it is a parallelogram.

Sizes of its sides: $PQ = SR = \dfrac{AC}{2}$ and $QR = PS = \dfrac{BD}{2}$, with $PQ \parallel AC$ and $QR \parallel BD$.
Its shape tells about the diagonals: $PQRS$ is a rhombus exactly when $AC = BD$; a rectangle exactly when $AC \perp BD$; a square exactly when the diagonals are both equal and perpendicular.
Its own diagonals: $PR$ and $QS$ bisect each other (they are the diagonals of a parallelogram). If $AC = BD$, then $PQRS$ is a rhombus, so $PR \perp QS$. The converse is also true.
Non-convex and self-intersecting quadrilaterals: the proof only uses the four triangles $ABC$, $ADC$, $BCD$, $BAD$, so the theorem still holds. In the special case where the lines $AC$ and $BD$ are parallel, the "parallelogram" collapses into a single segment.
Rebuilding $ABCD$ from $PQRS$: choose any point $A'$. Reflect it in $P$ to get $B'$, reflect $B'$ in $Q$ to get $C'$, reflect $C'$ in $R$ to get $D'$. Reflecting $D'$ in $S$ brings you back to $A'$ because $PQRS$ is a parallelogram. Then $A'B'C'D'$ is congruent to $ABCD$ (it has $P, Q, R, S$ as midpoints of its sides).

12. Applications of Parallelograms

Physics: a fundamental law uses parallelograms to combine quantities such as forces (studied in a later class).
Mechanical linkages: James Watt's parallel motion linkage, an important part of his steam engine, is based on parallelograms.
Pantograph: a parallelogram-based device that makes exact scaled copies of a drawing.
Inside geometry: parallelograms are a tool for proving facts about triangles. The Midpoint Theorem and the Centroid Theorem were both proved with their help.
Tiling: parallelograms tile the plane, and this leads to tilings with any triangle or quadrilateral.

13. Tiling the Plane with Any 4-gon

The key idea: around any vertex of a tiling, the angles of the tiles must add to $360^\circ$. The four angles of a quadrilateral also add to $360^\circ$. So four copies can fit around one point if each of the four angles is used once.
Several arrangements: there are multiple ways to place the four copies around a point, and any of them can be continued to cover the plane.

Method 1

Half-turn about a side's midpoint

Take two copies of the 4-gon. Rotate one by $180^\circ$ about the midpoint of a side. It fits exactly along that side. Repeat on every edge of every new copy.

Method 2

Varignon grid

The Varignon parallelograms of the tiles form a parallelogram grid. Start with that grid, place copies on alternate cells, and the gaps between them are again copies of the 4-gon.

Method 3

Slide along a diagonal

Two copies sharing a vertex are shifted copies of each other. Sliding a copy along a diagonal by the length of the diagonal gives the neighbour; repeating along both diagonals builds a grid of copies.

The half-turn sends $O$ to $M$ and $M$ to $O$, so the turned copy shares the whole side $OM$ with the original.

Why Method 1 works: a $180^\circ$ turn about the midpoint $N$ of $OM$ swaps $O$ and $M$, so the two copies match perfectly along $OM$. Where the copies meet at a vertex, all four angles occur once, adding to $360^\circ$. Any new copy can be obtained by turning any of its neighbours; both ways give the same result.
Method 2 details: in a tiling by SOME, the Varignon parallelograms of all the copies form the same grid as in the parallelogram tiling. Place a copy on each shaded cell using the reverse construction from the Varignon section. These copies meet only at vertices, and the gaps are again congruent 4-gons.
Non-convex 4-gons also tile: a non-convex quadrilateral like DART can be tiled by the same methods.
Triangles tile too: two copies of any triangle, with one turned $180^\circ$ about the midpoint of a side, form a parallelogram. Since parallelograms tile the plane, so does every triangle.
A regular pentagon cannot tile the plane: its interior angle is $108^\circ$, and $360^\circ$ is not a whole-number multiple of $108^\circ$ ($3 \times 108^\circ = 324^\circ$, $4 \times 108^\circ = 432^\circ$), so the angles at a vertex can never add to exactly $360^\circ$. Suitable irregular pentagons can tile; the most recent one was found in 2015.
A procedure needs justification: for each tiling method one must still prove that it covers the whole plane with no gaps and no overlaps.
Aperiodic tilings: some small sets of shapes tile the plane without ever repeating a pattern. They helped describe the atomic arrangement of quasicrystals, discovered by Dan Shechtman (Nobel Prize in Chemistry, 2011). The "hat" shape, found in 2023 by Smith, Myers, Kaplan and Goodman-Strauss, is the first single shape known to force such a tiling.

14. Extension Results (from the Exercises)

Result	Statement
Trapezium midline	In trapezium $ABCD$ with $AB \parallel DC$, let $E$ be the midpoint of $AD$. A line through $E$ parallel to $AB$ meets $BC$ at its midpoint $F$, and $EF = \dfrac{AB + CD}{2}$. Conversely, if $F$ is the midpoint of $BC$, then $EF \parallel AB$.
Trisecting a segment	In parallelogram $ABCD$, if $M$ and $N$ are the midpoints of $AB$ and $CD$, then $DM$ and $BN$ trisect the diagonal $AC$. This gives a method to trisect any segment.
Parallel line in a triangle	If $PQ \parallel BC$ with $P$ on $AB$ and $Q$ on $AC$, then $\dfrac{AP}{PB} = \dfrac{AQ}{QC}$ (proved here only when $\frac{AP}{PB}$ is rational; the general case comes in Grade 10).
Existence of a 4-gon with given sides	Four positive lengths are the sides of a 4-gon exactly when the longest is less than the sum of the other three. (For triangles: the sum of the two shorter sides must exceed the longest.)
Diagonals of an $n$-gon	An $n$-gon has $\dfrac{n(n-3)}{2}$ diagonals.
Angle sum of an $n$-gon	$(n - 2) \times 180^\circ$ for a planar, non-self-intersecting $n$-gon ($180^\circ$ for $n = 3$, $360^\circ$ for $n = 4$).

15. Summary and Quick Revision

No.	Theorem	Statement
1	Properties of a parallelogram	Opposite sides equal; opposite angles equal; diagonals bisect each other
2	Converse of (a)	Opposite sides equal $\Rightarrow$ parallelogram
3	Converse of (b)	Opposite angles equal $\Rightarrow$ parallelogram
4	Converse of (c)	Diagonals bisect each other $\Rightarrow$ parallelogram
5	Equal parallel sides test	One pair of opposite sides equal and parallel $\Rightarrow$ parallelogram
6	Midpoint Theorem	Segment joining midpoints of two sides is parallel to the third side and half of it
7	Converse of Midpoint Theorem	Line through a midpoint parallel to a second side bisects the third side
8	Centroid Theorem	Medians are concurrent; centroid divides each median $2 : 1$ from the vertex
9	Varignon's theorem	Midpoints of the sides of any 4-gon form a parallelogram

Strategy

Auxiliary construction

Draw an extra line or point (a parallel through a vertex, an extended segment) so that a parallelogram or a pair of congruent triangles appears.

Strategy

Run a proof backwards

To prove a converse, retrace the original proof in reverse and justify every step in the new direction.

Strategy

Break it down

Study one new segment or point at a time instead of many at once, as in the proofs of the Midpoint and Centroid Theorems.

Quick facts:
• A quadrilateral is convex when all internal angles are less than $180^\circ$, or equivalently when its diagonals intersect.
• Internal angles of a planar, non-self-intersecting 4-gon add to $360^\circ$.
• Every 4-gon (convex or non-convex) tiles the plane.

Common mistakes:
• Calling any four-sided figure a quadrilateral without checking collinearity, planarity and crossing sides.
• Naming vertices out of order (for example calling $ABCD$ by $ABDC$).
• Concluding "parallelogram" from one pair of equal sides, or from one pair of parallel sides, instead of both together.
• Believing that equal diagonals alone make a rectangle, or that perpendicular diagonals in a parallelogram make a square (it is a rhombus).
• Assuming a converse is true without proof.
• Using the Midpoint Theorem when only one of the two points is known to be a midpoint.
• Reversing the centroid ratio; the longer part ($2$) is toward the vertex.
• Forgetting that only one line passes through a given point parallel to a given line.
• Claiming that a regular pentagon tiles the plane.