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Adjacent and opposite elements

(iii) Definitions without vertex names: Opposite sides are two sides having no common endpoint. Opposite angles are internal angles at non-adjacent vertices; their vertices are joined by a diagonal, not by a side.

Define an internal angle precisely

Definition: At a vertex, the two rays along the adjacent sides divide a small neighbourhood into two angular sectors. The internal angle is the sector lying inside the bounded region enclosed by the quadrilateral. Choose the neighbourhood small enough that it meets no non-adjacent side.

At a convex vertex, this is the sector smaller than 180°. At a concave vertex, it is the sector greater than 180°. Thus we must not automatically choose the smaller angle between the two rays.

Using the opposite vertex: For the angle at A, consider triangle BCD. If A lies strictly inside triangle BCD, A is the reflex vertex and the internal angle is 360° minus the smaller angle between AB and AD. Otherwise A is convex and its internal angle is the smaller angle.

Which conditions force convexity?

(a) AB ∥ DC: no

The two opposite sides lie on distinct parallel lines. The other two sides connect their endpoints. For the boundary to be simple, the endpoints must be paired without crossing. The result is a convex trapezium: every turn is in the same direction and every interior angle is less than 180°. Reversing the pairing creates a crossing, not a concave simple quadrilateral.

(b) AB = CD: yes

Take A = (0, 0), B = (5, 0), C = (3, 1), D = (0, 5). Then AB = 5 and CD = √(3² + 4²) = 5. Point C lies inside triangle ABD, so ABCD is concave at C.

(c) ∠A = ∠C: yes

Take A = (−2, 0), B = (0, 1), C = (2, 0), D = (0, 4). Reflection in the vertical line interchanges A and C and fixes B and D, so ∠A = ∠C. But B lies inside triangle ACD and is a reflex vertex.

Classify the seven regions for D

A precise labelling method: Use A = (0, 0), B = (4, 0), C = (0, 3). For D = (x, y), put α = 1 − x/4 − y/3, β = x/4 and γ = y/3. The three boundary lines are α = 0, β = 0 and γ = 0. Also α + β + γ = 1, so all three cannot be negative.

These signs identify the seven regions without relying on a particular sketch. “Beyond a side” means across that side’s line from the opposite vertex.

How to check: Choose one point in a region and draw the boundary in the fixed order A → B → C → D → A. Moving D within the same region cannot change the type without crossing one of the three excluded lines.

Can a quadrilateral be crossed and non-planar?

Suppose opposite sides AB and CD cross at E. The two intersecting lines AB and CD determine one plane. Points A and B lie on the first line; C and D lie on the second. Hence all four vertices lie in that plane. The same argument applies if AD and BC cross.

True or false—with reasons

(i) A parallelogram with one right angle is a rectangle: TRUE

Opposite angles are equal and adjacent angles are supplementary. If one angle is 90°, each adjacent angle is 180° − 90° = 90°, and the opposite angle is also 90°. All four angles are right angles.

(ii) A rhombus with perpendicular diagonals is a square: FALSE

Every rhombus already has perpendicular diagonals. A rhombus with angles 60°, 120°, 60°, 120° satisfies the condition but is not a square. Add “one angle is 90°” or “the diagonals are equal” to force a square.

(iii) Equal diagonals in a parallelogram imply a rectangle: TRUE

Compare triangles ABC and BAD. AB is common, BC = AD (opposite sides), and AC = BD (given). Thus ΔABC ≅ ΔBAD by SSS. Hence ∠ABC = ∠BAD. These adjacent parallelogram angles also sum to 180°, so each is 90°. The parallelogram is a rectangle.

One diagonal bisects an angle

Given: ∠BAC = ∠CAD. Since AB ∥ CD, ∠BAC = ∠ACD. Since AD ∥ BC, ∠CAD = ∠BCA.

Therefore AC bisects ∠BCD. Also ∠BAC = ∠BCA, so triangle ABC is isosceles and AB = BC. Opposite sides of the parallelogram are equal: AB = CD and BC = AD.

Which converses are true?

(i) Must it be a rhombus? YES

Here the condition means AC bisects angles A and C, and BD bisects angles B and D. In triangles ABC and ADC, the angles at A are equal, the angles at C are equal and AC is common. By ASA, AB = AD and BC = CD. Similarly, triangles ABD and CBD are congruent using the bisected B and D angles and common BD; hence AB = BC. Combining gives all four sides equal.

(ii) Must it be a rhombus? YES

Let the diagonals meet at O. Bisection gives AO = CO and BO = DO, so ABCD is a parallelogram. Since AC ⟂ BD, triangles AOB and COB are congruent by SAS: AO = CO, OB is common and both included angles are 90°. Thus AB = CB. A parallelogram with equal adjacent sides is a rhombus.

(iii) Must it be a rectangle? NO

An isosceles trapezium can have equal diagonals without right angles. For example A = (−1, 1), B = (1, 1), C = (2, 0), D = (−2, 0) has AC = BD = √10, but is not a rectangle.

Extra condition: Require ABCD to be a parallelogram, or require the equal diagonals also to bisect each other. Then Exercise 12.2 Q1(iii) applies.

The angle-bisector rectangle

Let the bisectors of D and C meet at P, of C and B at Q, of B and A at R, and of A and D at S, as in the book. Adjacent angles of a parallelogram are supplementary.

In the triangle formed by A, B and the intersection of their bisector lines, the third angle is therefore 90°. Equivalently, the bisectors of adjacent supplementary angles are perpendicular. The same reasoning applies to every adjacent pair.

Consequently each of the four angles of PQRS is 90°. Opposite angle-bisector lines are parallel, so the four distinct intersections form a rectangle.

Why AB ≠ BC? If AB = BC, the parallelogram is a rhombus. Its internal angle bisectors are its diagonals: the A and C bisectors lie on one line, the B and D bisectors on the other. All four named intersections collapse to the diagonal intersection. There is no non-degenerate rectangle.

Four congruent triangles—and reconstruction

(i) Apply the Midpoint Theorem three times.

Each of the four triangles has side lengths AB/2, BC/2 and CA/2. Hence they are congruent by SSS. With corresponding vertices written in the correct order:

(ii) Construction

1. Through P draw a line parallel to QR.
2. Through Q draw a line parallel to PR.
3. Through R draw a line parallel to PQ.
4. Let the first two lines meet at A, the first and third at B, and the second and third at C.

These are the original side lines because APB ∥ QR, AQC ∥ PR and BRC ∥ PQ. Alternatively, the parallelograms created by the three parallel lines give AP = PB, AQ = QC and BR = RC. Thus P, Q, R really are the reconstructed side-midpoints.

The mid-segment bisects every segment from the vertex

Let X = MN ∩ AD. By the Midpoint Theorem in triangle ABC, MN ∥ BC. Since B, D and C are collinear, MX ∥ BD. In triangle ABD, M is the midpoint of AB and MX is parallel to BD. The converse Midpoint Theorem therefore gives AX = XD.

Join the midpoints of trapezium diagonals

Auxiliary point: Let E be the midpoint of AD. Apply the Midpoint Theorem in triangles ADC and ADB.

Since AB ∥ DC, EG and EH are the same line through E. Thus E, G and H are collinear, and that line is parallel to AB. Both G and H lie on the same ray from E for this simple trapezium.

If AB = CD, one pair of opposite sides is equal and parallel, so ABCD is a parallelogram. Its diagonals bisect each other, giving G = H. A single point does not determine a line GH.

Diagonals of the midpoint parallelogram

(i) By Varignon’s theorem, PQRS is a parallelogram. Therefore its diagonals PR and QS bisect each other.

(ii) Applying the triangle Midpoint Theorem gives:

If AC = BD, all four sides of PQRS are equal. It is a rhombus, whose diagonals PR and QS are perpendicular.

Converse: If PR ⟂ QS, the parallelogram PQRS is a rhombus. To see this, let its diagonals meet at O. OP = OR, OQ is common, and the included angles POQ and ROQ are right angles; hence ΔPOQ ≅ ΔROQ by SAS, so PQ = RQ. Therefore AC/2 = BD/2, giving AC = BD.

Recover a congruent quadrilateral from its midpoint frame

Important distinction: PQRS alone does not uniquely determine ABCD. The task allows you to refer to the given original ABCD while recreating a congruent copy. You must carry over the location of one original vertex, not choose it arbitrarily.

(i) A precise construction

1. Copy the parallelogram frame with corresponding points P′, Q′, R′, S′.
2. Copy triangle SPA onto side S′P′, preserving SA, PA and the original orientation. Its third vertex is A′. Choose the same side of S′P′ as in the original.
3. Reflect A′ in point P′ to get B′; that is, put A′, P′, B′ on one line with A′P′ = P′B′.
4. Reflect B′ in Q′ to get C′. Reflect C′ in R′ to get D′. Join the boundary.

(ii) Why it is congruent and closes

There is a rigid motion carrying the original labelled frame to the copied frame. The copied triangle SPA ensures that it also carries A to A′. Since P is the midpoint of AB, the image of B is exactly the point reflection of A′ in P′, namely B′. The same argument successively gives C′ and D′. Therefore all four new vertices are the images of the original vertices under the same rigid motion; the quadrilaterals are congruent. It also follows that S′ is the midpoint of D′A′, so the last side closes correctly.

An algebraic midpoint check is useful: B′ = 2P′ − A′, C′ = 2Q′ − B′ and D′ = 2R′ − C′. Because P′ + R′ = Q′ + S′ in a parallelogram, D′ = 2S′ − A′. This confirms the required collinearity and bisection on the final side. These equations are coordinate-wise midpoint formulas.

(iii) When is the frame a square?

If PQRS is a square, adjacent sides are equal and perpendicular; hence AC = BD and AC ⟂ BD. Conversely, if AC and BD are equal and perpendicular, PQRS has equal adjacent sides and a right angle. As it is already a parallelogram, it must be a square.

Why regular pentagons cannot tile the plane

At an edge-to-edge tiling vertex, a whole number of 108° angles would have to total 360°. But 360 ÷ 108 = 10/3, not an integer. Three angles give 324°, leaving a gap of 36°; four give 432°, causing overlap.

What if a vertex touches the middle of another tile’s edge? That edge contributes 180°. A local meeting would require 108k + 180l = 360 with integers k ≥ 1 and l ≥ 0. If l = 0, k = 10/3; if l = 1, k = 5/3; if l ≥ 2 there is no room for a 108° corner. None is possible.

Tile with a concave quadrilateral

One example has D = (0.45, 1), A = (1, 0), R = (0.45, 2), T = (0, 0), joined in that order. D lies inside triangle ART, so D is the reflex vertex.

1. Trace the quadrilateral. Mark the midpoint of each side.
2. At a boundary edge, rotate a copy by 180° about that edge’s midpoint. Its shared edge matches the original edge with its endpoints reversed.
3. Continue across exposed edges, identifying repeated copies when two construction routes arrive at the same location.
4. The repeating pattern contains translated copies in one orientation and half-turned copies in the other.

Alternative: Construct the Varignon midpoint grid first and place copies in the prescribed alternating positions; fill the remaining quadrilateral gaps with congruent half-turned copies. A proof that the repetitions have no gaps or overlaps is given in End Q23.

Tile the plane with any triangle

1. Take triangle ABC and mark the midpoint O of BC.
2. Rotate a second copy through 180° about O. A moves to D, B to C and C to B.
3. The two triangles form quadrilateral A–B–D–C. Its diagonals AD and BC bisect each other at O, so it is a parallelogram.
4. Tile the plane by translations of this parallelogram. Keep the dividing diagonal BC in every copy.

Each parallelogram is split into two congruent copies of the original triangle. Since the parallelograms tile with no gaps or overlaps, so do the triangles.

Find a midpoint using ruled paper

In the supplied figure, the segment’s endpoints are on horizontal lines separated by four equal gaps. The middle horizontal line is two gaps from each endpoint line. Its intersection with the slanting segment is the midpoint.

Justification: Draw a vertical segment from the upper endpoint to the horizontal line through the lower endpoint, forming a right triangle with the slanting segment. The middle horizontal line passes through the midpoint of this vertical side and is parallel to the horizontal base. By the converse Midpoint Theorem, it bisects the slanting side.

Angle sum of a crossed quadrilateral

Consider triangles AED and BEC. Their angles at E are vertically opposite; call each θ, with 0° < θ < 180°. The other angles are exactly the four smaller vertex angles of the crossed quadrilateral.

Make the sum 2°: Draw rays EA and ED making 179°. Choose EA = ED. Extend AE through E to B with EB = EA; extend DE through E to C with EC = ED. Then AED and BEC are congruent isosceles triangles, each with apex angle 179°. Each base angle is (180° − 179°)/2 = 0.5°.

Equal end-pieces on a diagonal

Let O = AC ∩ BD. The diagonals of ABCD bisect each other, so OA = OC and OB = OD. The condition DP = BQ means that P and Q are positioned symmetrically about the midpoint O of BD.

For the order shown in the book, OP = OD − DP and OQ = OB − BQ, which are equal. Thus O bisects both AC and PQ. These are the diagonals of APCQ. By the diagonal-bisection test, APCQ is a parallelogram.

A fold locates two midpoints

Let the crease meet AB at M and AC at N. Every point on the crease is equally distant from A and B, so the crease is the perpendicular bisector of AB. Therefore AM = MB and MN ⟂ AB.

Since BC ⟂ AB too, MN ∥ BC. In triangle ABC, the line through the midpoint M of AB parallel to BC meets AC at its midpoint by the converse Midpoint Theorem.

Cut and reassemble into congruent triangles

(i) Three congruent triangles from six median pieces

Let P, Q, R be midpoints of BC, CA, AB; let M be the centroid. Cut along the medians to obtain six small triangles. Pair the two pieces adjacent to each side.

For the BC pair, keep ΔMPC fixed and rotate ΔMPB by 180° about P. B lands on C because PB = PC. Let M land on M′. The shared sides PB and PC now match, while M, P, M′ lie on one straight line with P between M and M′. The two pieces form the single triangle MCM′.

Repeat at Q and R. Each assembled triangle has the same three side lengths: ⅔AP, ⅔BQ, ⅔CR. Therefore the three are congruent by SSS. The straight line MPM′ is the reason the assembled boundary is a triangle rather than a four-sided figure.

(ii) Repeat to obtain nine triangles

We show that the medians of triangle MCM′ have lengths AB/2, BC/2, AC/2. First, P is the midpoint of MM′, so its median from C has length CP = BC/2.

For the other two medians, a short coordinate argument makes the relationships explicit. Use M as origin; let the original vertex position vectors be a, b, c. The centroid property gives a + b + c = 0. Here P = −a/2 and M′ = −a, so the assembled triangle has vertices 0, c, −a.

Here |u − v| means the distance between the two represented points; the equations are just midpoint-coordinate calculations. Applying part (i) to each assembled triangle multiplies these median lengths by ⅔.

Each of the three triangles gives three more, producing nine congruent triangles.

(iii) Reassemble into two congruent triangles

1. Let P be the midpoint of BC. Cut ABC along median AP, leaving triangles APB and APC. Keep APC intact.
2. In triangle APB, let R be the midpoint of AB. Cut along its median PR, obtaining APR and BPR.
3. Rotate BPR through 180° about R. B lands on A; let P land on P′. Keep triangle APR fixed.
4. The rotated piece AP′R joins APR along AR. Since P, R, P′ are collinear with PR = RP′, the union is triangle APP′.

Quadrilateral A–P–B–P′ has diagonals AB and PP′ bisecting each other at R, so it is a parallelogram. Thus AP′ = BP = PC and P′A ∥ BP ∥ PC. Also P′A and PC have the same directed displacement, so A–P′–P–C is a parallelogram; hence PP′ = AC. Therefore triangles APP′ and APC have equal sides AP, PC and AC and are congruent by SSS.

The trapezium mid-segment theorem

(i) Parallel through a midpoint

Let M = EF ∩ BD. In triangle DAB, E is the midpoint of DA and EM ∥ AB. Thus M is the midpoint of DB and EM = AB/2. In triangle DBC, M is the midpoint of DB and MF ∥ DC. Thus F is the midpoint of BC and MF = DC/2.

(ii) Both leg-midpoints are given

Take M to be the midpoint of BD. In triangle DAB, EM ∥ AB. In triangle DCB, FM ∥ DC. Since AB ∥ DC, the unique line through M parallel to them contains both E and F. Hence E, M, F are collinear and EF ∥ AB. The same half-length relations again give the length formula.

A line through the diagonal intersection

Compare triangles AOP and COQ. We have AO = CO because the parallelogram diagonals bisect each other. Also ∠AOP = ∠COQ by vertically opposite angles, and ∠OAP = ∠OCQ because AB ∥ CD and AC is a transversal. Thus ΔAOP ≅ ΔCOQ by ASA.

Area ratio of two smaller trapezia

By End Q7, the mid-segment EF is parallel to the bases and:

The mid-segment is halfway between the parallel bases, so the two smaller trapezia have the same height, h/2, where h is the original height.

How many quadrilaterals do four points form?

(i) Count the cycles. Fix A as the starting vertex. The remaining three vertices can be arranged in 3! = 6 ways. Reversing the direction traces the same boundary, so divide by 2.

Representative orders are ABCD, ABDC and ACBD. Cyclic shifts or reversal do not produce a new quadrilateral.

Why? In convex position, only the order following the outside boundary is convex; each other order connects a crossing pair of diagonals as sides. If one point lies inside the other three, every boundary order has that point as a dent. Segments from the inner point stay within the outer triangle and cannot cross an unrelated outer edge.

Proportional division for rational ratios

Equal-intercept lemma: Equally spaced points on one transversal, with parallel lines drawn through them, give equally spaced intersection points on another transversal. To prove it for three consecutive parallels, call their first-transversal points U₀, U₁, U₂, with U₀U₁ = U₁U₂, and their second-transversal points V₀, V₁, V₂. Through V₀ draw a line parallel to U₀U₂, meeting the next two parallels at W₁, W₂. The parallelograms give V₀W₁ = W₁W₂. In triangle V₀W₂V₂, the line through midpoint W₁ parallel to W₂V₂ meets V₀V₂ at V₁, so V₀V₁ = V₁V₂ by the converse Midpoint Theorem. Applying this to consecutive triples proves the lemma.

(i) PB = 2

Divide PB into two pieces of length 1. AB is now split into three equal parts, the first being AP. Draw parallels to BC through the division points. The equal-intercept lemma splits AC into three equal parts; the first is AQ = 5. The remaining two make QC.

(ii) PB = 1/3

Divide AP into three pieces each of length 1/3. Together with PB, AB has four equal parts. The corresponding three parts of AQ total 5, so one part is 5/3; QC is one such part.

(iii) PB = 2/3

Divide AP into three pieces of length 1/3 and PB into two such pieces. On AC, AQ consists of three equal parts and QC of two. Each part is 5/3.

(iv) AP/PB is rational

Write AP/PB = m/n with positive integers m, n. There is a length u with AP = mu and PB = nu. Mark AB into m + n equal parts of length u and draw parallels to BC through these marks. The lemma gives equal intervals of length v on AC, with AQ = mv and QC = nv.

Trisect a diagonal and then any segment

(i) Prove the three parts equal

Since AM = MB = AB/2 and DN = NC = CD/2, we have MB = DN and MB ∥ DN. Thus M–B–N–D is a parallelogram, so DM ∥ BN. Let X = DM ∩ AC and Y = BN ∩ AC, with order A, X, Y, C.

In triangle ABY, M is the midpoint of AB and MX ∥ BY, so X is the midpoint of AY: AX = XY. In triangle CDX, N is the midpoint of CD and NY ∥ DX, so Y is the midpoint of CX: CY = YX.

(ii) Three methods for an arbitrary segment PQ

1. Parallelogram method: Mark the midpoint O of PQ. Choose B off line PQ and reflect B in O to get D. Then P–B–Q–D is a parallelogram. Mark M and N as midpoints of PB and QD. Draw DM and BN; their intersections with PQ trisect it by part (i).
2. Equal-step method: Draw an auxiliary ray from P. Mark U, V, W on it with PU = UV = VW. Join W to Q. Through U and V draw parallels to WQ to meet PQ. The equal-intercept lemma from Q11 makes the three portions of PQ equal.
3. Centroid method: Through Q draw a line different from PQ, and choose B, C on it with QB = QC. In triangle PBC, PQ is a median. Construct a second median to find centroid G. Then PG = 2PQ/3 and GQ = PQ/3. Bisect PG at H; now PH = HG = GQ.

Varignon beyond convex quadrilaterals

(i) Concave and crossed cases

Let P, Q, R, S bisect AB, BC, CD, DA. The midpoint theorem can still be applied separately to triangles ABC, ADC, BCD and BAD; it does not require the original four-sided boundary to be convex.

Thus the midpoint figure is a parallelogram whenever the two original diagonal directions are not parallel.

(ii) Degenerate exception

If AC ∥ BD, all four midpoint edges are parallel to one line and all four midpoints lie on that line. The midpoint figure collapses to a segment, possibly traced back and forth. For example A = (0, 0), B = (0, 1), C = (2, 0), D = (3, 1) gives midpoints (0, 0.5), (1, 0.5), (2.5, 0.5), (1.5, 0.5). This is a crossed original figure.

Conversely, if all midpoint sides lie on one line, PQ ∥ AC and QR ∥ BD force AC ∥ BD. For a genuine simple planar quadrilateral with no three collinear vertices, this degeneracy cannot occur.

(iii) Non-planar case

Yes. Each triangle used in the argument is planar even if the original four points are not. More explicitly, the coordinate-wise midpoint formulas give:

Therefore the midpoint of PR equals the midpoint of QS. Also S = P + R − Q, showing that S lies in the plane through P, Q, R. Hence the midpoint figure is a planar parallelogram even when ABCD is spatial. In the genuinely non-planar case the original diagonal lines cannot be parallel, since parallel lines would contain all four vertices in one plane.

Does an inscribed parallelogram force midpoints?

No. A counterexample is enough. Use the unit square A = (0, 0), B = (1, 0), C = (1, 1), D = (0, 1), and choose:

Each point lies one-quarter of the way along its side in the cyclic direction. None is a midpoint. Yet:

The diagonals PR and QS bisect each other, so PQRS is a parallelogram.

Alternative proofs of the midpoint theorem and converse

(i) P and Q are midpoints

Choose QS = PQ. Compare triangles APQ and CSQ. AQ = CQ because Q bisects AC, PQ = QS by construction, and ∠AQP = ∠CQS because they are vertically opposite. Thus ΔAPQ ≅ ΔCSQ by SAS.

Hence AP = CS and ∠APQ = ∠CSQ. These equal alternate angles give AP ∥ CS. Since A, P, B are collinear, PB ∥ CS, and PB = AP = CS. Therefore P–B–C–S is a parallelogram by the equal-and-parallel-side test.

(ii) P is midpoint and PQ ∥ BC

Extend the ray PQ to S so that PS = BC. Then PS ∥ BC and PS = BC, so P–B–C–S is a parallelogram. Consequently CS ∥ PB and CS = PB = PA. The line AC crosses PS at Q between P and S.

In triangles APQ and CSQ, ∠AQP = ∠CQS (vertically opposite), ∠APQ = ∠CSQ (alternate angles using AP ∥ CS), and AP = CS. Thus the triangles are congruent by AAS.

Converse statements for special quadrilaterals

This is an open exploration. The table covers the standard properties used in this chapter. A converse is a new statement: it needs a proof or a counterexample. Conditions in the “converse” column are the assumptions, not conclusions.

Angles of a non-planar quadrilateral

Proof using the hinge diagonal AC

Join AC. The triangles ABC and ADC each have angle sum 180°, so their six triangle angles total 360°. The angles at B and D are already the quadrilateral’s angles. At A, the angle between AB and AD is less than the sum of the angles made by those rays with AC when the two triangle planes are different. The same strict inequality holds at C.

This is the angle version of folding a hinge: with the two triangles flat on opposite sides of AC, the two angles add; folding out of that plane brings the outer rays closer. If either angle-sum exceeds 180°, the inequality is immediate because the smaller angle between two rays is at most 180°.

Construction with angle sum 2°

1. Construct an isosceles triangle ABC with AB = BC = 1 unit and apex angle ∠ABC = 0.5°. Call its short base length AC = d.
2. Make an identical triangle ADC with AD = DC = 1 and the same base AC. Treat AC as a hinge.
3. Rotate the second triangle about AC until BD = d. This is possible: BD varies continuously from 0 (coincident triangles) to twice their altitude (flat on opposite sides), and the small positive d lies strictly between these values.
4. Choose the intermediate, non-flat position with BD = d.

All four boundary sides are 1, and AC = BD = d. Each vertex angle is the apex angle of a triangle with side lengths 1, 1, d. By SSS, each is 0.5°. The hinge is neither fully closed nor fully open, so the four points are non-planar.

The diagonal-translation tiling method

(i) Four parallelograms identify the translation

Let the directed shift be from E to O. Set E′ = O and continue the line EO to O′ so that OO′ has the same length and direction as EO. Construct S′ and M′ so that E–S–S′–O and E–M–M′–O are parallelograms. These give:

The other two parallelograms are S–O–O′–S′ and M–O–O′–M′. Their opposite sides are equal and parallel because all four point displacements agree. Thus the four parallelograms are:

Every vertex is moved by one and the same translation, which preserves lengths and angles. Therefore the whole cutout matches S′O′M′E′ exactly, not just its diagonal.

(ii) Generate nine translated copies

Let u be the directed diagonal EO and v be the directed diagonal SM. Place copies translated by iu + jv for i, j each equal to −1, 0, 1. First add the four copies at ±u and ±v; then the four corner copies at ±u ±v. Together with the original, this gives nine.

These same-orientation copies meet only at vertices in the intended simple-quadrilateral pattern. Each gap is a half-turned copy of the original. Fill the gaps and continue the two translation directions. End Q23 proves that this process covers the plane without overlap.

Which side lengths are possible?

(i) Three sides

For a ≤ b ≤ c, draw a segment of length c. Draw circles of radii a and b at its endpoints. A non-collinear intersection exists exactly when |a − b| < c < a + b. Since c is the largest positive length, the first inequality is automatic. Thus the condition reduces to a + b > c. Equality produces a straight, degenerate figure.

(ii) Sides 2, 5, 11, x

Every side must be shorter than the sum of the other three. The two nontrivial restrictions are:

The other two inequalities hold automatically when x > 4.

(iii) General condition and construction

For positive a, b, c, d, a non-degenerate quadrilateral exists exactly when the longest length is less than the sum of the other three. Necessity follows because the three-side broken path joining a side’s endpoints is longer than the straight side.

For sufficiency, choose a diagonal length z satisfying both triangle inequalities:

The longest-side condition ensures these two open intervals overlap: |a − b| < c + d and |c − d| < a + b. Choose z in the overlap, avoiding any exceptional value that makes three vertices collinear. Construct triangles with sides (a, b, z) and (c, d, z) on opposite sides of their shared z-side. Their outer boundary is a simple quadrilateral with the required side lengths.

Count the diagonals of an n-gon

Definition: A diagonal is a segment joining two non-adjacent vertices. In a concave polygon a diagonal need not lie entirely inside the polygon.

Count from each vertex. It cannot be joined diagonally to itself or to either of its two neighbours, leaving n − 3 possible diagonal endpoints. Across n vertices this counts n(n − 3) endpoints, but each diagonal is counted twice, once from each end.

Add a new vertex between two neighbouring vertices A and B. The new vertex has diagonals to n − 2 of the old vertices (all except A and B). Also the former side AB becomes a diagonal. No old diagonal is lost.

Starting from D₃ = 0 gives 2, 5, 9, 14, 20, …, agreeing with the formula. The double-counting argument proves the formula for every n ≥ 3.

Angle sum of a simple planar polygon

Convex case: From one vertex, draw diagonals to every non-neighbouring vertex. They divide the polygon into n − 2 non-overlapping triangles. Every polygon angle is exactly the sum of the triangle angles at that vertex.

Concave case: Do not draw an arbitrary fan: some fan diagonals may leave the polygon. Instead triangulate using internal non-crossing diagonals. A simple polygon always has an internal diagonal that splits it into two smaller simple polygons; repeatedly cutting along such diagonals gives a triangulation. This is the standard triangulation fact for simple polygons.

To check the count independently, one such diagonal creates polygons with r and s vertices where r + s = n + 2, since its two endpoints are counted in both pieces. Inductively their angle sums are (r − 2)180° and (s − 2)180°. Adding:

Along the split, the two smaller angles reassemble into the original internal angle, including a reflex angle where applicable. The induction begins with triangles. Thus the formula remains valid for every simple planar polygon.

Multiple converses of the Midpoint Theorem

(i) PRLL + HALF ⇒ both midpoint conditions

Full statement: If P lies on AB and Q on AC of triangle ABC, PQ is parallel to BC and PQ has half the length of BC, then P and Q are the respective midpoints.

Let M and N be the actual midpoints of AB and AC. The Midpoint Theorem gives MN ∥ BC and MN = BC/2. Hence PQ ∥ MN and PQ = MN. If P ≠ M, the quadrilateral P–Q–N–M, lying between the two parallel segments, has one pair of opposite sides equal and parallel. It must be a parallelogram, forcing PM ∥ QN. But PM lies on AB and QN lies on AC; these side lines meet at A and cannot be parallel. This contradiction gives P = M. The unique parallel through P then gives Q = N.

(ii) The textbook’s converse

Keep the first sentence: “P is the midpoint of AB and Q lies on AC.” Reverse the implication in the next sentence: “If PQ ∥ BC, then Q is the midpoint of AC.” This is exactly Theorem 7. The half-length conclusion follows from the direct Midpoint Theorem.

(iii) The remaining type of pair: P MID + HALF

In words: If P is the midpoint of AB and Q lies on AC with PQ = BC/2, must Q be the midpoint and PQ be parallel to BC? No.

Take A = (0, 0), B = (2, 0), C = (3, 1), P = (1, 0), Q = (3/10, 1/10). Q lies one-tenth of the way from A to C, so it is not the midpoint.

Yet PQ is not parallel to BC: from P to Q the horizontal change is −7/10 and the vertical change is 1/10, whereas from B to C both changes are 1. Thus neither remaining conclusion follows.

There are six ways to choose two of four conditions, but swapping the labels P and Q makes two pairs equivalent. The direct theorem, (i), and Theorem 7 plus its symmetric version cover all except this last type and its symmetric version.

Why all three tiling methods really work

What must be proved? Angles adding to 360° at one vertex only gives a local check. A full proof must show that the repeated placements are consistent, cover arbitrarily distant points and never overlap in their interiors. The following common construction proves all three methods at once. This starred solution uses an optional slanted-coordinate argument; no distances are changed in the actual tiling.

1. Describe the two families of copies

Call the original simple quadrilateral A–B–C–D. Let u be the directed diagonal AC and v the directed diagonal BD. Place every translated copy T + iu + jv, where i and j are integers, and also every translated copy of T*, the half-turn of T about the midpoint of AB. Translations and half-turns preserve the shape, so every tile is congruent to T.

2. Choose convenient coordinates

The diagonal directions of a simple non-degenerate quadrilateral are not parallel. Choose their directions as the two coordinate axes and their lengths as coordinate units. Relabel cyclically if necessary so the line BD crosses the interior of segment AC. Then the coordinates can be written:

For a convex quadrilateral, −1 < t < 0. A concave one can have t outside that interval. The algebra below handles both. The axes may be oblique: we are only describing points and translations, not assuming that the diagonals are perpendicular.

3. Check a full vertical slice

Within 0 ≤ x ≤ 1 the original tile occupies exactly the interval between lower boundary L(x) and upper boundary U(x):

The half-turn sends (x, y) to (s − x, t − y). For 0 ≤ x ≤ s, use T* itself; for s ≤ x ≤ 1, use T* translated by (1, 0). Substituting into the boundary formulas shows that this companion tile occupies the interval:

The original occupies [L(x), U(x)]. Their interiors do not overlap, and together they fill the single interval [U(x) − 1, U(x)], of length 1. Translating these intervals by every integer in the vertical direction fills the entire vertical line with no interior overlaps or gaps. Translating horizontally by every integer then fills the entire plane. At the slice endpoints, tiles simply share boundary points or edges.

4. Identify each textbook procedure with these copies

What fraction of the square is shaded?

Step 1 — Choose a convenient side length. A ratio of areas is unchanged by scaling, so take the outer square’s side as 2. Label it as in the diagram: A = (0, 2), B = (2, 2), C = (2, 0), D = (0, 0). Then E = (1, 2), F = (2, 1), G = (1, 0), H = (0, 1).

Step 2 — Write equations of the four drawn lines. These follow directly from their two labelled endpoints.

Step 3 — Find two adjacent shaded vertices. P is AG ∩ HB:

Q is EC ∩ HB:

Step 4 — Establish the shaded figure is a square. AG ∥ EC and HB ∥ DF, so PQRS is a parallelogram. A line with horizontal/vertical changes (1, −2) is perpendicular to one with changes (2, 1), since a quarter-turn sends the first direction to the second. Thus the two line families are perpendicular. The original construction is invariant under a 90° rotation about the outer square’s centre, so the central parallelogram has four equal sides. Hence it is a square.

Step 5 — Use Pythagoras for its side and area. From P to Q, the horizontal change is 4/5 and the vertical change is 2/5.

What counts as a quadrilateral?

Can every ordinary quadrilateral tile? Yes: every simple planar quadrilateral, including a concave one, admits the half-turn/translation tiling proved in End Q23.

Definition: Four distinct coplanar vertices, with no three collinear, are joined in cyclic order by four straight segments. Adjacent sides meet only at their shared endpoint and non-adjacent sides do not intersect. This describes a simple planar quadrilateral; a concave one is allowed.

Careful reading: The chapter’s Definition 1 box should be read together with its preceding “no three vertices collinear” restriction, which is also included in the chapter summary. Without that restriction, a four-point boundary with a straight angle could slip into the wording.

Convexity: All four interior angles are below 180° in a convex quadrilateral. Its two diagonal segments intersect internally. A simple concave quadrilateral has exactly one reflex angle and its diagonal segments do not intersect.

Why the parallelogram converses work

Opposite sides equal

If AB = CD and BC = DA, join AC. Triangles ABC and CDA have three corresponding equal sides, including AC common, so they are congruent by SSS. Corresponding angles along AC are equal, giving AB ∥ CD and BC ∥ DA.

Opposite angles equal

Every adjacent pair is supplementary. Using the converse of the co-interior-angle property gives both pairs of opposite sides parallel. Conversely, if every adjacent pair is supplementary, subtracting A + B = 180° and B + C = 180° gives A = C, and similarly B = D. Thus the blank is: “adds up to 180°”.

Diagonals bisect each other

At their intersection O, AO = CO and BO = DO. Triangles AOB and COD are congruent by SAS using the vertically opposite included angles. Equal alternate angles give AB ∥ CD. Similarly, triangles AOD and COB give AD ∥ BC.

One pair equal and parallel

If AB ∥ DC and AB = DC, compare triangles BAC and DCA: BA = DC, AC is common, and ∠BAC = ∠DCA by alternate angles. SAS gives ∠BCA = ∠DAC, so BC ∥ AD. Both pairs are parallel.

Midpoints and four congruent pieces

Initially the midpoint marks give only AP = PB, AQ = QC, BR = RC. These alone do not immediately provide all three corresponding sides of the four small triangles. First establish the Midpoint Theorem using an auxiliary parallelogram, as in End Q15. It supplies PQ = BC/2, QR = AB/2 and PR = AC/2. SSS then proves the four small triangles congruent.

A parallel to BC through the midpoint P of AB passes through the midpoint Q of AC. One short proof is to mark the actual midpoint N of AC: PN ∥ BC by the direct theorem. Since there is only one parallel through P to BC, PN and PQ are the same line, so N = Q.

For a point P not necessarily a midpoint, the expected generalisation is AP/PB = AQ/QC. The rational-ratio case was proved in End Q11; the unrestricted theorem is developed in Grade 10.

A second proof of the Centroid Theorem

Choose S so that M is the midpoint of AS: extend AM beyond M by MS = AM. In triangle ABS, P and M are midpoints of AB and AS. Therefore PM ∥ BS and BS = 2PM. Since P, M, C lie on one median, BS ∥ MC.

Similarly, in triangle ACS, Q and M are midpoints, so QM ∥ CS and CS = 2QM. Because B, M, Q are collinear, CS ∥ BM. Hence B–S–C–M is a parallelogram.

Let X = SM ∩ BC. Its diagonals SM and BC bisect each other, so X is the midpoint of BC. Since A, M, S are collinear, AX is the third median and passes through M.

From local fitting to a full tiling

For a simple quadrilateral, the internal angles sum to 360°, including a reflex angle if present. Thus a suitable arrangement with one copy of each angle can fill one complete turn around a vertex. This explains the local fit but does not by itself prove a global tiling.

A half-turn about a side-midpoint reverses that side’s endpoints and gives its exact neighbouring tile. Two consecutive half-turns act as a translation. In the repeating pattern, those translations are the directed diagonals of the original quadrilateral. Adding a horizontal-grid shift and then a vertical-grid shift gives the same final displacement as doing them in the opposite order; this accounts for agreeing construction routes.

The Varignon frame has sides parallel to the original diagonals with half their lengths. It therefore provides the same two-direction grid for organising the copies. Place a congruent tile consistently around each prescribed frame; do not assume the frame alone determines the tile. End Q23 supplies the full no-gap/no-overlap proof.