Class 9 Mathematics Chapter 12: Quadrilaterals

The four vertices must be distinct points lying in one plane. No three of them may be collinear. Every point of the figure other than the four vertices must lie on exactly one of the four segments.

A quadrilateral is convex when every one of its internal angles is less than $180^\circ$, that is, when it has no dent. An equivalent test is that its two diagonals cross each other; in a non-convex quadrilateral they do not.

A diagonal joins two vertices not already joined by a side. Each vertex has exactly one such partner, so there are $4 \times 1 = 4$ vertex-to-partner joins, and each segment gets counted from both ends. That gives $4 \div 2 = 2$ diagonals.

Call the middle one of the three collinear points its vertex. The two sides meeting there both lie along the one line, so they do not turn a corner. The figure therefore encloses either a triangular region or nothing at all, and the requirement that no three vertices be collinear is exactly what rules this out.

Every quadrilateral has internal angles totalling $360^\circ$. Two angles of $190^\circ$ already come to $380^\circ$, which overshoots the total before the other two angles are counted at all. Since an internal angle cannot be zero or negative, such a figure is impossible, and this is why at most one dent can occur.

ABCD, BCDA, CDAB, DABC going one way round, and DCBA, ADCB, BADC, CBAD going the other way: eight in all. Any other order would call for a diagonal to be a side.

The sides adjacent to PQ are SP and QR, since each shares a vertex with it. The side opposite PQ is RS, which shares no vertex with PQ.

The three given angles total $100 + 110 + 125 = 335^\circ$, so the fourth is $360 - 335 = 25^\circ$. Every angle is now below $180^\circ$, so the quadrilateral is convex.

At the crossing there is a point of the figure which is not a vertex and which lies on two different sides. The definition permits such a point on exactly one side, so the figure is excluded. When we want to admit it we call it a self-intersecting quadrilateral, and these do arise in practice, for instance in the folded positions of a four-bar linkage.

Four points joined in order need not lie in a common plane, and the resulting non-planar 4-gons are perfectly real figures. But the properties studied here, such as internal angles totalling $360^\circ$ and diagonals that cross, are statements about figures drawn in one plane, so the definition requires all four vertices to be coplanar.

The first is not. Two of its sides lie along one line, so there is no corner there and no four-sided region is enclosed.

The second is not, under the ordinary definition. The crossing point is a non-vertex point lying on two sides, whereas the definition allows exactly one. If we choose to admit it we call it self-intersecting.

The third IS a quadrilateral. Nothing in the definition limits angle size; the large angle simply makes it non-convex.

The fourth is not, for this chapter. It is a non-planar 4-gon, and the definition requires all four vertices to lie in one plane.

Draw a diagonal of the quadrilateral. It cuts the figure into two triangles, and the angles of each triangle total $180^\circ$. Between them those six angles make up the four internal angles of the quadrilateral exactly once each, so the four total $2 \times 180^\circ = 360^\circ$.

A dent at a vertex means the internal angle there is greater than $180^\circ$. Two dents would contribute more than $360^\circ$ on their own, leaving a negative amount for the remaining two angles, which is impossible. So at most one internal angle can exceed $180^\circ$ and a quadrilateral has at most one dent.

The first test reads the angles: a quadrilateral is convex when all four internal angles are less than $180^\circ$. The second reads the diagonals: a quadrilateral is convex when its two diagonals cross each other.

Take a quadrilateral with angles $85^\circ$, $95^\circ$, $100^\circ$ and $80^\circ$, which total $360^\circ$. Every angle is below $180^\circ$, so the angle test calls it convex, and its diagonals are found to meet at an interior point, so the diagonal test agrees.

Now push one vertex inward until its internal angle becomes $210^\circ$, with the other three at $70^\circ$, $60^\circ$ and $20^\circ$. The angle test now reports non-convex, and the diagonals are seen to miss each other, so again the two tests agree. The second test is often quicker in a drawing, since crossing is easier to see than measuring four angles.

The pins hold the four strips at fixed lengths, so the side lengths never change. What changes are the angles. Pushing the frame leaves it a closed planar four-sided figure throughout, so it is a quadrilateral at every stage, but its angles move continuously.

As the push continues the frame passes through a convex stage, then reaches a position where one internal angle grows past $180^\circ$ and the figure becomes non-convex, and if pushed further two of the strips can come to lie across each other, giving a self-intersecting configuration.

An engineer designing such a linkage must predict where the bars go in every one of these positions, not just the tidy convex ones. That is why it is worth defining non-convex and self-intersecting 4-gons precisely instead of dismissing them as not being quadrilaterals.

The opposite sides of a parallelogram are equal in length; the opposite angles are equal; and the diagonals bisect each other.

Property form: if a quadrilateral is a parallelogram, its opposite sides come out equal. Converse form: a quadrilateral with equal opposite sides must be a parallelogram.

Both pairs of opposite sides are parallel, which is the definition. Both pairs of opposite sides are equal. Both pairs of opposite angles are equal. The diagonals bisect each other. One pair of opposite sides is both equal and parallel.

$\angle C = 65^\circ$, since opposite angles are equal. $\angle B$ and $\angle D$ are each adjacent to $\angle A$, so each is $180^\circ - 65^\circ = 115^\circ$.

Opposite sides are equal, so $CD = 9$ cm and $DA = 6$ cm. The perimeter is $9 + 6 + 9 + 6 = 30$ cm, which is $2 \times (9 + 6)$.

The diagonals bisect each other, so the crossing point is the midpoint of both. Along the longer diagonal the two distances are $14 \div 2 = 7$ cm each, and along the shorter they are $10 \div 2 = 5$ cm each.

An isosceles trapezium has diagonals of equal length but only one pair of parallel sides, so it is not a parallelogram. A single counterexample is enough to defeat the claim. What does characterise a parallelogram is the diagonals BISECTING each other, which is a different condition from being equal in length.

The two conditions have to apply to the SAME pair of sides. An isosceles trapezium has one pair of parallel sides and the OTHER pair equal, and it is not a parallelogram. The correct test, the equal parallel sides test, requires one pair to be both equal and parallel at once.

A side joining them acts as a transversal across the two parallel opposite sides. The two angles are then interior angles on the same side of that transversal, so they total $180^\circ$.

Any two adjacent angles total $180^\circ$, and an internal angle must be greater than $0^\circ$. So no single angle can reach $180^\circ$, which means none can exceed it either. With every angle below $180^\circ$ the figure has no dent.

Let $ABCD$ be a parallelogram, so $AB \parallel DC$ and $AD \parallel BC$. Draw the diagonal $AC$.

$AC$ is a transversal for the parallel pair $AB$ and $DC$, so the alternate angles $\angle BAC$ and $\angle DCA$ are equal. $AC$ is also a transversal for the parallel pair $AD$ and $BC$, so $\angle DAC$ and $\angle BCA$ are equal. The side $AC$ is common to both triangles.

Comparing triangle $ACD$ with triangle $CAB$, we have two pairs of equal angles with the included side $AC$ common, so the triangles are congruent by ASA.

Corresponding sides of congruent triangles are equal, which gives $AB = DC$ and $AD = BC$. The opposite sides are therefore equal.

Let $ABCD$ be a quadrilateral with $AB = DC$ and $AD = BC$. Draw the diagonal $AC$.

In triangle $ACD$ and triangle $CAB$ we have $AD = CB$, $DC = BA$, and $AC$ common. All three pairs of sides are equal, so the triangles are congruent by SSS.

Corresponding angles are therefore equal, so $\angle DCA = \angle BAC$. These are alternate angles for the transversal $AC$ crossing $DC$ and $AB$, and equal alternate angles mean $AB \parallel DC$. Likewise $\angle DAC = \angle BCA$ gives $AD \parallel BC$. Both pairs of opposite sides are parallel, so $ABCD$ is a parallelogram.

The two proofs travel the same route in opposite directions: both turn on the congruence of the same two triangles. But they justify that congruence differently, by ASA going forwards and by SSS coming back, and they end at different conclusions. That is why the converse needs its own proof and cannot simply cite the property.

Let $ABCD$ be a quadrilateral whose diagonals $AC$ and $BD$ meet at $E$, with $E$ the midpoint of each. So $EA = EC$ and $EB = ED$.

Compare triangle $AED$ with triangle $CEB$. We have $EA = EC$ and $ED = EB$ by hypothesis, and the included angles $\angle AED$ and $\angle CEB$ are vertically opposite, hence equal. The triangles are congruent by SAS.

Therefore $\angle DAE = \angle BCE$. These are alternate angles for the transversal $AC$ crossing $AD$ and $BC$, so $AD \parallel BC$.

Comparing triangle $AEB$ with triangle $CED$ in the same way gives $\angle BAE = \angle DCE$, and hence $AB \parallel DC$.

Both pairs of opposite sides are parallel, so $ABCD$ is a parallelogram. Note that the congruence here is justified by SAS, where the forward property used ASA or AAS — again a separate argument for a separate claim.

Let $ABCD$ be a quadrilateral with $AB \parallel DC$ and $AB = DC$. Let the diagonals meet at $E$.

$AC$ is a transversal for the parallel pair $AB$ and $DC$, so $\angle BAE = \angle DCE$. Similarly $BD$ is a transversal for the same pair, so $\angle ABE = \angle CDE$. We also have $AB = DC$ by hypothesis.

Comparing triangle $EAB$ with triangle $ECD$, two pairs of angles are equal and the sides between them are equal, so the triangles are congruent by ASA. Hence $EA = EC$ and $EB = ED$, which says the diagonals bisect each other. By the test already proved for bisecting diagonals, $ABCD$ is a parallelogram.

The word "both" is essential because the two conditions must fall on the SAME pair of sides. A quadrilateral with one pair of sides parallel and the OTHER pair equal need not be a parallelogram at all: an isosceles trapezium is exactly such a figure, with one pair of parallel sides and the remaining two sides equal, and it has only one pair of parallel sides.

Join the midpoints of any two sides of a triangle. The segment you obtain runs parallel to the remaining side, and its length is half of that side's.

Take the midpoint of one side of a triangle and draw a line through it parallel to a second side. Such a line must cut the remaining side exactly in half, and the segment it cuts off is half the side it parallels.

The three medians of a triangle pass through one common point, called the centroid. That point divides each median in the ratio $2 : 1$, with the longer part joining the centroid to the vertex.

By the Midpoint Theorem that segment is half the third side, so it measures $18 \div 2 = 9$ cm.

Each midpoint segment is half the side it is parallel to, so the new triangle has sides $10 \div 2 = 5$ cm, $14 \div 2 = 7$ cm and $18 \div 2 = 9$ cm. Its perimeter is $5 + 7 + 9 = 21$ cm. The original perimeter is $10 + 14 + 18 = 42$ cm, so the new perimeter is exactly half.

The centroid divides a median in the ratio $2 : 1$, so the median consists of three equal parts of $24 \div 3 = 8$ cm. The part from the centroid to the vertex is two of them, $16$ cm, and the part from the centroid to the midpoint of the side is $8$ cm.

The four midpoints of a quadrilateral's sides form the corners of a parallelogram, known as its Varignon parallelogram. It applies to every quadrilateral, including non-convex ones.

Each side of the Varignon parallelogram joins the midpoints of two sides of a triangle cut off by a diagonal, so by the Midpoint Theorem it is half that diagonal. The sides are therefore $16 \div 2 = 8$ cm and $22 \div 2 = 11$ cm, each appearing twice, giving a perimeter of $2 \times (8 + 11) = 38$ cm. Note that this is just $16 + 22$, the sum of the diagonals.

The four angles of a quadrilateral total $360^\circ$, which is exactly the turn needed to close up around a point. So four copies can be fitted round a common point with each angle used once, leaving no gap. Carrying this out everywhere, by rotating each new copy through $180^\circ$ about the midpoint of a shared edge, covers the plane without gaps or overlaps.

Copies meeting at a point must have angles totalling $360^\circ$. Each angle of a regular pentagon is $108^\circ$, and $108^\circ$ does not divide $360^\circ$ exactly: three give $324^\circ$, which leaves a gap, and four give $432^\circ$, which overlaps.

Let $P$ and $Q$ be the midpoints of sides $AB$ and $AC$ of triangle $ABC$. We must show $PQ \parallel BC$ and $PQ = \frac{1}{2} BC$.

Construct the line through $C$ parallel to $BA$, and let it meet the line $PQ$ at $R$.

In triangles $APQ$ and $CRQ$ we have $AQ = CQ$, since $Q$ is the midpoint of $AC$; $\angle AQP = \angle CQR$, as they are vertically opposite; and $\angle APQ = \angle CRQ$, because $AP$ and $RC$ are parallel by construction. The triangles are therefore congruent by AAS.

From the congruence, $PQ = QR$, so $PQ = \frac{1}{2} PR$, and $CR = AP$. Since $P$ is the midpoint of $AB$ we have $AP = BP$, and so $CR = BP$.

Now $BP$ and $CR$ are parallel by construction and equal in length, so by the equal parallel sides test $BCRP$ is a parallelogram. Hence $PQ \parallel BC$, and $PQ = \frac{1}{2} PR = \frac{1}{2} BC$ because $PR = BC$ as opposite sides of the parallelogram.

Let $P$, $Q$, $R$ and $S$ be the midpoints of the sides $AB$, $BC$, $CD$ and $DA$ of quadrilateral $ABCD$. Draw the diagonal $AC$.

In triangle $ABC$, the points $P$ and $Q$ are midpoints of two sides, so by the Midpoint Theorem $PQ \parallel AC$ and $PQ = \frac{1}{2} AC$. In triangle $ADC$, the points $S$ and $R$ are midpoints of two sides, so $SR \parallel AC$ and $SR = \frac{1}{2} AC$.

Both $PQ$ and $SR$ are therefore parallel to $AC$, hence parallel to each other, and both are half of $AC$, hence equal. One pair of opposite sides of $PQRS$ is both equal and parallel, so by the equal parallel sides test $PQRS$ is a parallelogram.

For the perimeter, apply the same reasoning to the other diagonal $BD$: the sides $QR$ and $PS$ are each parallel to $BD$ and equal to $\frac{1}{2} BD$. So the four sides of $PQRS$ are $\frac{1}{2} AC$, $\frac{1}{2} BD$, $\frac{1}{2} AC$ and $\frac{1}{2} BD$, and the perimeter is $2 \left( \frac{1}{2} AC + \frac{1}{2} BD \right) = AC + BD$. The perimeter of the Varignon parallelogram is exactly the sum of the two diagonals of the original quadrilateral.

Nothing in this argument assumed the quadrilateral was convex, so the result holds for a dented quadrilateral too.

Let $P$, $Q$ and $R$ be the midpoints of the sides $AB$, $AC$ and $BC$ of triangle $ABC$, and join them.

By the Midpoint Theorem applied three times: $PQ = \frac{1}{2} BC$, $QR = \frac{1}{2} AB$ and $PR = \frac{1}{2} AC$.

Consider the corner triangle at $A$, namely triangle $APQ$. Its sides are $AP = \frac{1}{2} AB$, $AQ = \frac{1}{2} AC$ and $PQ = \frac{1}{2} BC$. The central triangle $PQR$ has sides $QR = \frac{1}{2} AB$, $PR = \frac{1}{2} AC$ and $PQ = \frac{1}{2} BC$. The two triangles have the same three side lengths, so they are congruent by SSS.

The same comparison works for the corner triangle at $B$, which has sides $BP = \frac{1}{2} AB$, $BR = \frac{1}{2} BC$ and $PR = \frac{1}{2} AC$, and for the corner triangle at $C$. So all four small triangles are congruent to one another.

Each small triangle has all three sides exactly half those of $ABC$, so each is a half-size copy of the original triangle, with the same three angles. Since the four congruent pieces together make up $ABC$, each has one quarter of its area.

Method. Start with one copy of the quadrilateral. Take a second copy laid exactly on top of it, and rotate that copy through $180^\circ$ about the midpoint of one of the edges. The rotated copy comes to rest alongside the original, meeting it exactly along that shared edge. Now repeat this at every free edge of every copy placed so far, continuing outwards without limit.

Why it works. A half-turn about the midpoint of an edge carries that edge onto itself, exchanging its two endpoints, so the new copy matches the original along the whole edge with neither gap nor overlap.

At each vertex of the resulting pattern, four copies come together, and the half-turns arrange matters so that each of the four angles of the quadrilateral appears exactly once. Those four angles total $360^\circ$, which is precisely the full turn around a point, so the copies close up perfectly with nothing left over and nothing overlapping.

Since this works at every edge and every vertex, and no quadrilateral was assumed to be special, the plane can be tiled with copies of ANY quadrilateral, convex or not.

The four vertices must be distinct points lying in one plane. No three of them may be collinear. Every point of the figure other than the four vertices must lie on exactly one of the four segments.

A quadrilateral is convex when every one of its internal angles is less than $180^\circ$, that is, when it has no dent. An equivalent test is that its two diagonals cross each other; in a non-convex quadrilateral they do not.

A diagonal joins two vertices not already joined by a side. Each vertex has exactly one such partner, so there are $4 \times 1 = 4$ vertex-to-partner joins, and each segment gets counted from both ends. That gives $4 \div 2 = 2$ diagonals.

The opposite sides of a parallelogram are equal in length; the opposite angles are equal; and the diagonals bisect each other.

Property form: if a quadrilateral is a parallelogram, its opposite sides come out equal. Converse form: a quadrilateral with equal opposite sides must be a parallelogram.

Both pairs of opposite sides are parallel, which is the definition. Both pairs of opposite sides are equal. Both pairs of opposite angles are equal. The diagonals bisect each other. One pair of opposite sides is both equal and parallel.

Join the midpoints of any two sides of a triangle. The segment you obtain runs parallel to the remaining side, and its length is half of that side's.

Take the midpoint of one side of a triangle and draw a line through it parallel to a second side. Such a line must cut the remaining side exactly in half, and the segment it cuts off is half the side it parallels.

The three medians of a triangle pass through one common point, called the centroid. That point divides each median in the ratio $2 : 1$, with the longer part joining the centroid to the vertex.