Chapter 12: Quadrilaterals (Quadrilaterals)

(i) Adjacent to AB: BC and DA (each shares a vertex with AB). Opposite to AB: CD only.

(ii) Adjacent to ∠A: ∠B and ∠D (at the vertices joined to A by a side). Opposite to ∠A: ∠C only.

(iii) Two sides of a quadrilateral are opposite if they have no common endpoint. Two internal angles are opposite if their vertices are not the two ends of a side (that is, the two vertices are joined by a diagonal).

(i) Adjacent: BC, DA; opposite: CD. (ii) Adjacent: ∠B, ∠D; opposite: ∠C. (iii) Opposite sides share no vertex; opposite angles are at vertices joined by a diagonal.

Take the vertex A of quadrilateral ABCD; its adjacent vertices are B and D and its opposite vertex is C. The two rays AB and AD make an angle ∠BAD that is less than 180° (the usual angle), and also a reflex angle $360°-\angle BAD$.

Definition: if the opposite vertex C lies inside the angle ∠BAD (the part of the plane between the rays AB and AD), the internal angle at A is ∠BAD. Otherwise the internal angle at A is the reflex angle $360°-\angle BAD$.

For a convex quadrilateral the opposite vertex always lies inside ∠BAD, so every internal angle is the usual one. At the "dent" of a non-convex quadrilateral (such as D in DART), the opposite vertex lies outside the small angle, so the internal angle is reflex (more than 180°).

The internal angle at A is ∠BAD if the opposite vertex C lies inside ∠BAD, and 360° − ∠BAD (reflex) otherwise.

AB ∥ DC: No. A and B lie on one line and C and D on a parallel line, so the quadrilateral lies in the strip between the two lines. At A, side AB runs along the edge of the strip and side AD goes into the strip, so the internal angle at A is less than 180°; the same holds at B, C and D. So ABCD must be convex. (This is also why the diagonals of such a quadrilateral always intersect.)

AB = CD: Yes, it can be non-convex. Example: A(0, 0), B(4, 0), C(1, 1), D(1, 5). Here AB = 4 = CD, and the internal angle at C is reflex.

∠A = ∠C: Yes. Example: an arrowhead (dart) that is symmetric about BD, with A(−2, 0), B(0, 1), C(2, 0), D(0, 4). By symmetry ∠A = ∠C, and B is a dent with a reflex angle.

With AB ∥ DC it must be convex. With AB = CD, or with ∠A = ∠C, it can be non-convex.

The three lines cut the plane into 7 regions: the inside of △ABC, three regions across the sides (across AB, BC and CA), and three regions at the vertices (beyond A, beyond B, beyond C). Remember that ABCD has sides AB, BC, CD, DA and diagonals AC, BD.

• D inside △ABC: non-convex (the dent is at D).
• D across side CA (on the other side of AC from B, between the lines AB and BC): convex (the diagonals AC and BD cross).
• D across side AB: self-intersecting (side CD crosses side AB).
• D across side BC: self-intersecting (side DA crosses side BC).
• D in the region beyond vertex A, B or C: non-convex (the dent is at A, B or C respectively).

Convex: 1 region (across CA). Self-intersecting: 2 regions (across AB and across BC). Non-convex: the other 4 regions (inside the triangle and the three vertex regions).

No. "Self-intersecting" means two sides, say AB and CD, cross at a point E. Two lines that meet at a point lie in one plane, so the lines AB and CD lie in a single plane, and therefore all four points A, B, C, D lie in it. So the quadrilateral is planar.

No. If two sides cross, the two crossing lines (and hence all four vertices) lie in one plane.

(i) True. Adjacent angles of a parallelogram add up to 180°, so if one angle is 90°, its neighbours are 90° and the opposite angle is also 90°. All angles are right angles.

(ii) False. Every rhombus has perpendicular diagonals, including a rhombus with angles 60° and 120°, which is not a square.

(iii) True. Let ABCD be a parallelogram with AC = BD. In △ABC and △DCB: AB = DC (opposite sides), BC is common, and AC = DB. So △ABC ≅ △DCB (SSS), giving ∠ABC = ∠DCB. These are adjacent angles, so they add up to 180°; hence each is 90°, and by (i) ABCD is a rectangle.

(i) True (ii) False (iii) True

We are given ∠DAC = ∠CAB.

• AB ∥ DC with transversal AC gives ∠CAB = ∠ACD (alternate angles).
• AD ∥ BC with transversal AC gives ∠DAC = ∠ACB (alternate angles).

Hence ∠ACD = ∠CAB = ∠DAC = ∠ACB, so AC bisects ∠C.

Also ∠DAC = ∠DCA, so △ADC is isosceles with AD = DC. A parallelogram with two adjacent sides equal has all four sides equal (opposite sides are equal), so ABCD is a rhombus.

Alternate angles give ∠ACD = ∠CAB = ∠DAC = ∠ACB, so AC bisects ∠C; and ∠DAC = ∠DCA gives AD = DC, so ABCD is a rhombus.

(i) Yes. Since AC bisects ∠A and ∠C: in △ABC and △ADC, AC is common, ∠BAC = ∠DAC and ∠BCA = ∠DCA, so △ABC ≅ △ADC (ASA), giving AB = AD and CB = CD. In the same way, BD bisecting ∠B and ∠D gives BA = BC and DA = DC. So all four sides are equal.

(ii) Yes. Diagonals that bisect each other make ABCD a parallelogram (Theorem 4). If they meet at E at right angles, then △AEB ≅ △AED (SAS: EB = ED, AE common, right angles), so AB = AD. A parallelogram with adjacent sides equal is a rhombus.

(iii) No. An isosceles trapezium (or a suitable kite) has equal diagonals but is not a rectangle. Extra condition: if the equal diagonals also bisect each other, then ABCD is a parallelogram with equal diagonals, hence a rectangle (Q1 (iii)).

(i) Yes (ii) Yes (iii) No; it becomes Yes if the diagonals also bisect each other.

Adjacent angles of a parallelogram add up to 180°, e.g. ∠A + ∠B = 180°. The bisectors of ∠A and ∠B meet at a point, say S, forming △ASB with

$$\angle SAB+\angle SBA=\frac{1}{2}\angle A+\frac{1}{2}\angle B=90°\Rightarrow \angle ASB=90°$$

The same reasoning at each pair of adjacent vertices shows that the bisectors of ∠B and ∠C, of ∠C and ∠D, and of ∠D and ∠A also meet at right angles. So the four points where the bisectors meet form a quadrilateral PQRS whose four angles are all 90°: a rectangle.

Why AB ≠ BC: if AB = BC, ABCD is a rhombus, whose angle bisectors are its diagonals. All four bisectors then pass through the single point where the diagonals meet, and the "rectangle" shrinks to a point.

Each pair of adjacent bisectors meets at 90° (half of 180°), so PQRS has four right angles. If AB = BC, all four bisectors pass through one point and there is no rectangle.

(i) By the Midpoint Theorem:

$$PQ=\frac{BC}{2},QR=\frac{AB}{2},PR=\frac{AC}{2}$$

Each of the four small triangles has sides equal to half the sides of △ABC:

• △QPA: QP = BC/2, PA = AB/2, AQ = AC/2
• △PBR: PB = AB/2, BR = BC/2, PR = AC/2
• △QRC: QC = AC/2, RC = BC/2, QR = AB/2
• △PQR: PQ = BC/2, QR = AB/2, PR = AC/2

So by SSS, △PQR ≅ △QPA ≅ △PBR ≅ △QRC (with the vertices matched as the equal sides show).

(ii) Yes. The Midpoint Theorem says each side of △ABC is parallel to a side of △PQR: AB ∥ QR (and passes through P), AC ∥ PR (through Q), BC ∥ PQ (through R). So through each vertex of △PQR draw the line parallel to the opposite side. The three lines meet in pairs at A, B and C, and form the original triangle.

(i) △PQR ≅ △QPA ≅ △PBR ≅ △QRC (all have sides AB/2, BC/2, CA/2). (ii) Yes: draw through P, Q, R lines parallel to QR, PR, PQ; they form △ABC.

By the Midpoint Theorem, MN ∥ BC. Let MN meet AD at X. In △ABD, M is the midpoint of AB and MX ∥ BD (since MN ∥ BC). By Theorem 7, the line through the midpoint of one side parallel to another side bisects the third side, so X is the midpoint of AD.

MN ∥ BC, so in △ABD the line MX through the midpoint M is parallel to BD and therefore bisects AD (Theorem 7).

Let M be the midpoint of AD.

• In △ADC: M and G are midpoints of AD and AC, so MG ∥ DC.
• In △DAB: M and H are midpoints of DA and DB, so MH ∥ AB ∥ DC.

MG and MH both pass through M and are parallel to DC. There is only one line through M parallel to DC, so M, G and H lie on this line. Hence GH ∥ DC ∥ AB.

If AB = CD, then ABCD has a pair of equal and parallel opposite sides, so it is a parallelogram (Theorem 5), and its diagonals bisect each other: G = H. Then there is no segment GH to talk about.

MG ∥ DC and MH ∥ AB ∥ DC through the same point M, so G, H, M are collinear and GH ∥ AB. If AB = CD, ABCD is a parallelogram and G = H.

(i) By Theorem 9, PQRS is a parallelogram. PR and QS are its diagonals, and the diagonals of a parallelogram bisect each other.

(ii) By the Midpoint Theorem, $PQ=\frac{1}{2}AC$ and $QR=\frac{1}{2}BD$. If AC = BD, then PQ = QR, so the parallelogram PQRS has equal adjacent sides: it is a rhombus. The diagonals of a rhombus are perpendicular, so PR ⊥ QS.

Converse: yes. If PR ⊥ QS, the parallelogram PQRS has perpendicular diagonals, so it is a rhombus (Ex. 12.2 Q3 (ii)). Then PQ = QR, i.e. $\frac{1}{2}AC=\frac{1}{2}BD$, so AC = BD.

(i) They are the diagonals of the Varignon parallelogram. (ii) AC = BD makes PQRS a rhombus, so PR ⊥ QS; the converse is also true.

(i) Many quadrilaterals have the same Varignon parallelogram, so we also need to know where one vertex goes. Copy PQRS, and place A′ in the same position relative to P and S as A is in the original (for example, using the lengths AP and AS). Then:

• B′ is the point on line A′P beyond P with PB′ = A′P (P is the midpoint of A′B′);
• C′ is on line B′Q beyond Q with QC′ = B′Q;
• D′ is on line C′R beyond R with RD′ = C′R.

(ii) Each step is a half-turn (180° rotation) about a midpoint. A half-turn about P followed by a half-turn about Q moves every point by the fixed amount twice PQ (in the direction P to Q); the half-turns about R and S together move it by twice RS. Since PQRS is a parallelogram, $\overrightarrow{RS}=-\overrightarrow{PQ}$, so the four half-turns bring D′ back so that S is exactly the midpoint of D′A′: the figure closes up, and S lies on D′A′.

Because A′ was placed like A, and PB′ = PA′ with A′, P, B′ in a line (as for A, P, B), △A′PS ≅ △APS and the same reasoning step by step (triangles such as △SD′R ≅ △SDR) shows every vertex is placed exactly like the original. So A′B′C′D′ ≅ ABCD.

(iii) By the Midpoint Theorem, PQ ∥ AC with $PQ=\frac{1}{2}AC$, and QR ∥ BD with $QR=\frac{1}{2}BD$.

• If PQRS is a square, then PQ = QR, so AC = BD; and PQ ⊥ QR, so AC ⊥ BD.
• Conversely, if AC = BD and AC ⊥ BD, then PQ = QR and PQ ⊥ QR. A parallelogram with two equal adjacent sides meeting at a right angle is a square.

(i) Place A′ as in the original, then reflect it successively through P, Q, R to get B′, C′, D′. (ii) The four half-turns close up because PQRS is a parallelogram, and each step matches the original. (iii) PQ = AC/2 ∥ AC and QR = BD/2 ∥ BD, so "PQRS is a square" ⟺ "AC = BD and AC ⊥ BD".

Each interior angle of a regular pentagon is $\frac{(5-2)\times 180°}{5}=108°$. In a tiling, the angles that meet around any point must add up to 360°.

• If only corners of pentagons meet at a point: we need $108°\times k=360°$, so $k=\frac{10}{3}$, not a whole number.
• If the point lies in the middle of a side of one pentagon (a straight angle of 180°) and corners of others meet there: we need $180°+108°\times k=360°$, so $k=\frac{5}{3}$, again not a whole number.

So copies of a regular pentagon can never fit around a point without gaps or overlaps.

The angle 108° does not divide 360° (or 180°) a whole number of times, so the pentagons cannot fit around a vertex.

Method 1 (half-turns): take DART and turn a copy through 180° about the midpoint of one side. The copy fits exactly against that side. Repeat this at every side of every new copy. The four angles of DART meet at every vertex (each once), so they add up to 360°, and the tiles fit without gaps or overlaps, even though DART has a reflex angle.

In the figure, the hatched dart is the starting tile; the shaded tiles are its half-turned copies. Notice that every other tile is just a sliding copy of the starting tile, moved along one of its diagonals.

Method 2 (Varignon grid): draw the Varignon parallelogram of DART and tile the plane with it. Place copies of DART on alternate parallelograms of the grid, as in the text; the gaps left over are again copies of DART.

Method 1 is usually easier to carry out with paper cut-outs, because each new piece is found from its neighbour by a single half-turn.

Rotate copies of DART through 180° about the midpoints of its sides again and again; the pieces fill the plane (see the figure).

Take △ABC and a copy turned through 180° about the midpoint M of BC. The copy has the same sides, and AB and CA′ are equal and parallel (they are swapped by the half-turn), so ABA′C is a quadrilateral with a pair of equal parallel opposite sides: a parallelogram (Theorem 5). We know the plane can be tiled with copies of any parallelogram (the sliding grid of the introduction). Each parallelogram is made of two copies of the triangle, so the plane is tiled by the triangle.

Two copies of the triangle (one turned about the midpoint of a side) form a parallelogram, and parallelograms tile the plane.

In Fig. 12.33 the segment runs from one ruled line to another four lines up, crossing three lines in between. Its midpoint is where it crosses the middle one of these lines (the second line from either end).

Justification: equally spaced parallel lines cut every line across them into equal parts. Take any two consecutive gaps: the segment crosses three consecutive ruled lines ℓ₁, ℓ₂, ℓ₃ at X, Z, Y. From X drop a perpendicular to ℓ₃, meeting it at W. Since the lines are equally spaced, ℓ₂ passes through the midpoint of XW, and ℓ₂ is parallel to WY (both lie along the ruling). So in △XWY, by Theorem 7, ℓ₂ bisects XY: XZ = ZY. Applying this to each pair of consecutive gaps, the four crossings divide the segment into 4 equal parts, so the middle crossing point is its midpoint.

(If the segment spans an odd number of gaps, its midpoint lies halfway between the two middle lines.)

The midpoint is where the segment crosses the middle ruled line, because equally spaced parallel lines divide the segment into equal parts (Theorem 7).

The sides AB and CD cross at E, so the figure is made of two triangles, △EAD and △EBC, with vertically opposite angles at E: $\angle AED=\angle BEC=\theta$. The angles of the quadrilateral are angles of these triangles: ∠A = ∠DAE, ∠D = ∠ADE, ∠B = ∠EBC, ∠C = ∠BCE. So

$$\angle A+\angle B+\angle C+\angle D=(180°-\theta )+(180°-\theta )=360°-2\theta <360°$$

Sum = 2°: we need $2\theta =358°$, i.e. θ = 179°. Take A(−1, 0), B(1, 0) on a line through E = (0, 0), and D, C on a second line through E at 1° to the first, with $\angle AED=179°$: D = (cos 1°, sin 1°), C = (−cos 1°, −sin 1°). Then ABCD is a very flat "bow-tie" with angle sum 2°.

The sum is 360° − 2θ, where θ is the angle at the crossing; it equals 2° when θ = 179° (a very flat bow-tie).

Let O be the point where the diagonals AC and BD meet. In a parallelogram the diagonals bisect each other: OA = OC and OB = OD. Then

$$OP=OD-DP=OB-BQ=OQ$$

So the diagonals AC and PQ of quadrilateral APCQ bisect each other at O, and by Theorem 4, APCQ is a parallelogram.

OA = OC and OP = OD − DP = OB − BQ = OQ, so the diagonals of APCQ bisect each other.

In Fig. 12.35 the right angle is at B, with AB along one side and BC along the other. When A is folded onto B, the crease is the perpendicular bisector of AB: it passes through the midpoint P of AB and is perpendicular to AB.

Since BC is also perpendicular to AB, the crease is parallel to BC. So the crease is the line through the midpoint P of AB parallel to BC, and by Theorem 7 it bisects the third side AC. The point where the crease meets AC is the midpoint of AC.

The crease is the perpendicular bisector of AB; since BC ⊥ AB too, the crease is parallel to BC through the midpoint of AB, so it also passes through the midpoint of AC (Theorem 7).

(i) Take the two pieces △MPB and △MPC on either side of PM. Since PB = PC, we can turn △MPB about P so that PB lies along PC, with B landing on C and M landing at a point M′. Then ∠M′PC = ∠MPB, and $\angle MPC+\angle MPB=180°$ (B, P, C are collinear), so M, P, M′ lie on a straight line. The two pieces therefore form a triangle MCM′, with P the midpoint of MM′.

Its sides (using the Centroid Theorem, M divides each median 2 : 1):

$$M{M}^{\prime }=2MP=\frac{2}{3}AP,MC=\frac{2}{3}CR,M^{\prime }C=MB=\frac{2}{3}BQ$$

The same construction with the other two pairs of pieces (around Q and around R) gives triangles whose sides are again $\frac{2}{3}$ of the three medians. So we get 3 congruent triangles (SSS), each with sides equal to two-thirds of the medians of △ABC.

(ii) Apply (i) to each new triangle: its pieces reassemble into triangles with sides $\frac{2}{3}$ of its medians. One can check (using the length of a median, $m_a^2=\frac{2b^2+2c^2-a^2}{4}$) that a triangle whose sides are the medians $m_a,m_b,m_c$ of △ABC has medians $\frac{3}{4}a,\frac{3}{4}b,\frac{3}{4}c$. Our triangle has sides $\frac{2}{3}$ of the medians, so its medians are $\frac{2}{3}\times \frac{3}{4}$ of a, b, c, i.e. $\frac{a}{2},\frac{b}{2},\frac{c}{2}$. Two-thirds of these are $\frac{a}{3},\frac{b}{3},\frac{c}{3}$. So the 9 triangles have sides $\frac{AB}{3},\frac{BC}{3},\frac{AC}{3}$.

(iii) Yes. The median AP cuts △ABC into △APB and △APC, which have equal areas (equal bases BP = PC, same height) but are not congruent in general. Now cut △APB along the median from P (to the midpoint N of AB) and turn the piece △PNB through 180° about N, so that B goes to A. The two pieces of △APB now form a triangle with the same base and height arrangement as △APC, and, as in Chapter 6 (End-of-Chapter Q20), triangles on equal bases with the same height can be cut and rearranged into each other. In this way △ABC is cut into pieces that form 2 congruent triangles.

(i) Pieces on either side of each midpoint form a triangle with sides (2/3) of the three medians, giving 3 congruent triangles. (ii) Repeating gives 9 triangles with sides AB/3, BC/3, CA/3. (iii) Yes, starting with median AP and a median of △APB.

(i) Draw the diagonal AC and let EF meet it at G.

• In △ADC: E is the midpoint of AD and EG ∥ DC, so G is the midpoint of AC and $EG=\frac{DC}{2}$ (Theorem 7).
• In △CAB: G is the midpoint of CA and GF ∥ AB, so F is the midpoint of CB and $GF=\frac{AB}{2}$.

$$EF=EG+GF=\frac{AB+CD}{2}$$

(ii) Draw through E the line parallel to AB; by (i) it meets BC at its midpoint. But F is the midpoint of BC, and there is only one midpoint, so this line passes through F. Hence EF ∥ AB.

(The other way, with the midpoint M of BD: EM ∥ AB in △DAB and MF ∥ DC ∥ AB in △BDC, so E, M, F are on one line parallel to AB. The first way is shorter.)

(i) Using diagonal AC twice with Theorem 7: F is the midpoint and EF = EG + GF = (AB + CD)/2. (ii) The line through E parallel to AB meets BC at its midpoint, which is F.

In △OAP and △OCQ:

• OA = OC (diagonals of a parallelogram bisect each other)
• ∠OAP = ∠OCQ (alternate angles, AB ∥ DC with transversal AC)
• ∠AOP = ∠COQ (vertically opposite angles)

So △OAP ≅ △OCQ (ASA), and OP = OQ.

△OAP ≅ △OCQ (ASA), so OP = OQ and O is the midpoint of PQ.

By Q7, EF ∥ AD ∥ BC and

$$EF=\frac{AD+BC}{2}=\frac{3+5}{2}=4\text{cm}$$

Since E is the midpoint of AB, the line EF is halfway between AD and BC, so both smaller trapeziums have height $\frac{h}{2}$, where h is the height of ABCD.

$$\text{area}(AEFD)=\frac{1}{2}(3+4)\cdot \frac{h}{2}=\frac{7h}{4},\text{area}(EBCF)=\frac{1}{2}(4+5)\cdot \frac{h}{2}=\frac{9h}{4}$$

area(AEFD) : area(EBCF) = 7 : 9

(i) A quadrilateral is a closed path through the four points. There are $4\times 3\times 2\times 1=24$ orders of the letters, but each quadrilateral has 8 names (start at any of the 4 vertices and go either way round, e.g. ABCD = BCDA = ... = DCBA). So there are $\frac{24}{8}=3$ quadrilaterals: ABCD, ABDC and ACBD.

(ii) It depends on the position of the points.

• If the four points form a convex shape (none inside the triangle of the other three): one of the three quadrilaterals is convex (the one going round the outside), and the other two are self-intersecting (they use a diagonal as a side).
• If one point lies inside the triangle formed by the other three: none is self-intersecting and none is convex; all three are non-convex (the inside point is the dent).

(i) 3 quadrilaterals. (ii) Points in convex position: 1 convex and 2 self-intersecting. One point inside the triangle of the others: 0 convex, 0 self-intersecting (all 3 non-convex).

Key fact (from Theorem 7 and Q7): if parallel lines cut equal segments on one line, they cut equal segments on any other line crossing them.

(i) Divide AB into 3 equal parts of length 1 (AP = 1, PB = 2). The lines through the division points parallel to BC cut AC into 3 equal parts, each equal to AQ = √5. So $QC=2\sqrt{5}$.

(ii) Divide AP into 3 equal parts of length $\frac{1}{3}$; then PB is one more such part, so AB has 4 equal parts. The parallels cut AC into 4 equal parts, and AQ (3 parts) $=\sqrt{5}$, so each part is $\frac{\sqrt{5}}{3}$. QC is 1 part: $QC=\frac{\sqrt{5}}{3}$.

(iii) Divide AB into parts of length $\frac{1}{3}$: AP is 3 parts and PB is 2 parts. AQ (3 parts) = √5, so each part of AC is $\frac{\sqrt{5}}{3}$ and $QC=\frac{2\sqrt{5}}{3}$.

(iv) If $\frac{AP}{PB}=\frac{m}{n}$ (m, n natural numbers), there is a length t with AP = mt and PB = nt. Mark points dividing AB into m + n parts of length t and draw lines through them parallel to BC. They cut AC into m + n equal parts, say of length s, with m of them in AQ and n in QC. So

$$\frac{AQ}{QC}=\frac{ms}{ns}=\frac{m}{n}=\frac{AP}{PB}$$

(i) QC = 2√5 (ii) QC = √5/3 (iii) QC = 2√5/3 (iv) Split AB into equal parts; the parallels split AC into the same number of equal parts, so AQ/QC = AP/PB.

(i) MB = $\frac{1}{2}AB=\frac{1}{2}DC$ = DN and MB ∥ DN, so MBND is a parallelogram (Theorem 5), and DM ∥ NB. Let DM and BN meet AC at X and Y.

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

So AX = XY = YC: the segments DM and BN trisect AC.

(ii) Three ways to trisect a segment PQ:

• Using (i): choose any point B not on PQ and construct the parallelogram with diagonal PQ: let O be the midpoint of PQ and take D on BO extended with OD = BO. Then P, B, Q, D form a parallelogram with AC = PQ. Join D to the midpoint of PB and B to the midpoint of QD; these lines trisect PQ.
• Using Exercise 11: draw any ray from P and mark three equal steps P₁, P₂, P₃ on it with a compass. Join P₃Q, and draw lines through P₁ and P₂ parallel to P₃Q. They cut PQ into three equal parts.
• Using the Centroid Theorem: choose any point X not on line PQ and let Y be on XQ extended with QY = XQ, so PQ is a median of △PXY. Draw a second median (from Y to the midpoint of PX); it meets PQ at the centroid G, with PG = $\frac{2}{3}$PQ. Then G and the midpoint of PG are the two trisection points.

(i) MBND is a parallelogram, and two uses of Theorem 7 give AX = XY = YC. (ii) Parallelogram method, equal steps with parallels, or the centroid of a triangle having PQ as a median.

(i) Yes, in both cases. The proof only uses the Midpoint Theorem in the four triangles ABC, ADC, BCD, BAD. These triangles exist whatever the shape of ABCD, so PQ ∥ AC ∥ SR and QR ∥ BD ∥ PS still hold, and PQRS is a parallelogram.

(ii) PQ is parallel to AC and QR is parallel to BD. If the diagonals AC and BD are parallel (which can happen for a self-intersecting "quadrilateral", e.g. A(0, 0), B(0, 1), C(2, 0), D(2, 1)), then PQ and QR lie on the same line, and the four midpoints are collinear: the parallelogram flattens into a segment.

(iii) Yes. Each of the four triangles is still a flat triangle, so the Midpoint Theorem applies in it: PQ ∥ AC ∥ SR with PQ = SR = AC/2, and similarly for QR and PS. Two parallel lines lie in one plane, so P, Q, R, S are coplanar and form a parallelogram even when A, B, C, D do not lie in one plane.

(i) True in all cases; the proof only uses triangles. (ii) When AC ∥ BD (then the midpoints are collinear). (iii) Yes, PQRS is a flat parallelogram even for a non-planar ABCD.

No. Counterexample: let ABCD be a parallelogram with centre O (where its diagonals meet). Take any P on AB and let R be the point on CD with CR = AP; take any Q on BC and let S be the point on DA with DS = BQ. The half-turn about O swaps A ↔ C, B ↔ D, and so swaps P ↔ R and Q ↔ S. Hence O is the midpoint of both PR and QS: the diagonals of PQRS bisect each other, so PQRS is a parallelogram (Theorem 4). But P and Q can be chosen anywhere, not just at the midpoints.

For example, in a square of side 4, take AP = 1, BQ = 3, CR = 1, DS = 3.

No. In a parallelogram ABCD, any P, Q with R and S placed symmetrically (CR = AP, DS = BQ) give a parallelogram PQRS.

(i) Extend PQ to S with QS = PQ. In △APQ and △CSQ: AQ = QC (Q is the midpoint), PQ = SQ (by construction), and ∠AQP = ∠CQS (vertically opposite). So △APQ ≅ △CSQ (SAS). Hence:

• CS = AP = PB.
• ∠QCS = ∠QAP, which are alternate angles for lines CS and AB with transversal AC, so CS ∥ AB, i.e. CS ∥ PB.

So PBCS has a pair of equal and parallel opposite sides and is a parallelogram (Theorem 5). Therefore PS ∥ BC and PS = BC, so PQ ∥ BC and $PQ=\frac{1}{2}PS=\frac{1}{2}BC$.

(ii) Now P is the midpoint of AB and PQ ∥ BC, with Q on AC. Extend PQ to S so that PS = BC. Then PS is equal and parallel to BC, so PBCS is a parallelogram: CS = PB = PA and CS ∥ AB. In △APQ and △CSQ: AP = CS, ∠QAP = ∠QCS (alternate angles) and ∠APQ = ∠CSQ (alternate angles, AB ∥ CS with transversal PS). So △APQ ≅ △CSQ (ASA), giving AQ = QC (Q is the midpoint of AC) and PQ = QS, i.e. $PQ=\frac{1}{2}PS=\frac{1}{2}BC$.

(i) Take QS = PQ; then △APQ ≅ △CSQ (SAS), PBCS is a parallelogram, and PQ = ½BC ∥ BC. (ii) Take PS = BC; then PBCS is a parallelogram and △APQ ≅ △CSQ (ASA), so Q is the midpoint of AC.

Here are the main properties, their converses (for a quadrilateral that is not self-intersecting) and whether each converse is true.

Several converses are true (all sides equal ⇒ rhombus; diagonals bisecting each other at right angles ⇒ rhombus; equal bisecting diagonals ⇒ rectangle; equal, perpendicular, bisecting diagonals ⇒ square), while "equal diagonals ⇒ rectangle" and "perpendicular diagonals ⇒ rhombus" are false.

Split ABCD along the diagonal AC into △ABC and △ADC, which lie in two different planes. At vertex A, the angle of the quadrilateral is ∠DAB, while the triangles have angles ∠CAB and ∠DAC there. When the three rays AB, AC, AD do not lie in one plane, the angle between AB and AD is less than the sum of the other two:

$$\angle DAB<\angle DAC+\angle CAB$$

(Think of a corner of a box: the angles between the three edges at a corner satisfy a "triangle inequality".) In the same way, $\angle BCD<\angle BCA+\angle ACD$. Adding,

$$\angle A+\angle B+\angle C+\angle D<(\angle CAB+\angle ABC+\angle BCA)+(\angle DAC+\angle ACD+\angle CDA)=180°+180°=360°$$

Sum 2°: take two congruent, very thin triangles ABC and ADC with ∠B = ∠D = 0.9°. When they are flat and on opposite sides of AC, the sum is 360°. Rotate △ADC about the hinge AC towards △ABC: the angles at A and C shrink, and when D comes down onto B the angles at A and C become 0, leaving a sum of $0.9°+0.9°=1.8°$. The sum changes continuously during the rotation, so at some position in between it is exactly 2°.

At A and C, the angle of the quadrilateral is less than the sum of the two triangle angles there, so the total is less than 2 × 180°. Folding two thin congruent triangles about AC can make the sum exactly 2°.

(i) Sliding the cut-out along the line EO by the length EO moves every point by the same distance in the same direction. For each vertex X of SOME and its new position X′, the segment XX′ is equal and parallel to EO = E′O′. So SS′, OO′, MM′ and EE′ are all equal and parallel, and the four quadrilaterals SS′O′O, OO′M′M, MM′E′E and EE′S′S each have a pair of equal parallel opposite sides: they are parallelograms (Theorem 5). In a parallelogram the other two sides are also equal and parallel, so S′O′ = SO, O′M′ = OM, M′E′ = ME and E′S′ = ES, with the same directions. Hence S′O′M′E′ is an exact (shifted) copy of SOME.

(ii) Sliding along both diagonals, forwards and backwards, places copies at all positions SOME + (whole multiples of the two diagonal shifts). These form a 3 by 3 (and then larger) grid of copies that touch only at vertices. Each blank space is surrounded by four copies; turning SOME through 180° about the midpoint of a side (Method 1) shows that each blank space is exactly a half-turned copy of SOME. Together these give the tiling.

(i) The slide makes SS′, OO′, MM′, EE′ equal and parallel, giving four parallelograms, so each side of the copy equals and is parallel to the matching side of SOME. (ii) Sliding along both diagonals produces the grid, and the gaps are half-turned copies of SOME.

(i) Draw a segment of length c. Draw circles of radii a and b about its two ends. The circles meet (giving the third vertex) exactly when a + b > c; if a + b ≤ c, they do not meet off the line, and no triangle is possible.

(ii) Rule for a 4-gon: the longest side must be less than the sum of the other three (going straight is the shortest path between two vertices).

• 100: is 100 < 2 + 5 + 11 = 18? No. Not possible.
• 10: the longest side is 11, and 11 < 2 + 5 + 10 = 17. Possible.
• 1: is 11 < 2 + 5 + 1 = 8? No. Not possible.

For a fourth side x: if x ≥ 11 we need x < 2 + 5 + 11 = 18; if x < 11 we need 11 < 2 + 5 + x, i.e. x > 4. So the possible lengths are 4 < x < 18.

(iii) A 4-gon with sides a, b, c, d exists exactly when the largest of the four numbers is less than the sum of the other three. Necessity: one side is a straight segment, which is shorter than the path along the other three sides. Sufficiency: if the condition holds, we can choose the length of a diagonal so that both triangles it creates satisfy the triangle condition of (i), and then build the 4-gon from the two triangles.

(ii) 100: no; 10: yes; 1: no; possible lengths are those between 4 and 18 (4 < x < 18). (iii) A 4-gon exists exactly when the longest side is less than the sum of the other three.

(i) A diagonal of an n-gon is a segment joining two vertices that are not adjacent (not the ends of one side).

(ii) Formula: $\frac{n(n-3)}{2}$.

Justification: from each vertex we can draw a diagonal to every vertex except itself and its two neighbours, i.e. to n − 3 vertices. That gives n(n − 3) diagonals, but each diagonal has been counted twice (once from each end), so the number is $\frac{n(n-3)}{2}$.

Adding a vertex: going from n to n + 1 sides adds n − 1 diagonals (the new vertex joins n − 2 non-neighbours, and the old side that is "cut" becomes one more diagonal). Check: 5 → 6 adds 4 (5 + 4 = 9), 6 → 7 adds 5 (9 + 5 = 14). Indeed $\frac{(n+1)(n-2)}{2}-\frac{n(n-3)}{2}=n-1$.

An n-gon has $\frac{n(n-3)}{2}$ diagonals (5-gon: 5, 6-gon: 9, 7-gon: 14, 8-gon: 20); adding a side adds n − 1 diagonals.

Values: 5-gon 540°, 6-gon 720°, 7-gon 900°, 8-gon 1080°.

Formula: $(n-2)\times 180°$.

Proof: an n-gon can be cut by n − 3 non-crossing diagonals into n − 2 triangles (for a convex polygon, draw all diagonals from one vertex; every polygon, even a non-convex one, can be cut up into n − 2 triangles in this way, since one can always find a diagonal lying inside the polygon). The angles of these triangles together make up exactly the angles of the polygon, so the sum is $(n-2)\times 180°$.

The sum of the angles of an n-gon is (n − 2) × 180°.

(i) Statement: In △ABC, let P be on AB and Q on AC. If PQ ∥ BC and PQ = BC/2, then P and Q are the midpoints of AB and AC.

Proof: through Q draw the line parallel to AB, meeting BC at R. Then PBRQ has both pairs of opposite sides parallel: it is a parallelogram, so BR = PQ = BC/2. Hence R is the midpoint of BC. In △CAB, the line through the midpoint R of CB parallel to BA meets CA at Q, so by Theorem 7, Q is the midpoint of CA. Now in △ABC, the line through the midpoint Q of AC parallel to CB meets AB at P, so by Theorem 7 again, P is the midpoint of AB.

(ii) Converse of the second sentence: "If PQ ∥ BC, then Q is the midpoint of AC" (keeping "P is the midpoint of AB, Q on AC"). This is exactly Theorem 7. In named conditions: if (P MID) and (PRLL), then (Q MID) (and (HALF)).

(iii) Statement: "In △ABC, let P be the midpoint of AB and Q a point on AC with PQ = BC/2. Then Q is the midpoint of AC and PQ ∥ BC." This is false in general. The points Q on AC with PQ = BC/2 lie on a circle with centre P and radius BC/2, which can cut AC in two points. One is the midpoint N of AC; the other is not.

Example: A(0, 0), B(4, 0), C(3, 3). P = (2, 0) and BC/2 = √10/2 ≈ 1.58. The point Q′ = (0.5, 0.5) on AC has PQ′ = √(2.25 + 0.25) ≈ 1.58 = BC/2, but Q′ is not the midpoint (1.5, 1.5) of AC, and PQ′ is not parallel to BC.

(i) True: build parallelogram PBRQ and use Theorem 7 twice. (ii) Theorem 7: (P MID) and (PRLL) ⇒ (Q MID) and (HALF). (iii) Not necessarily: e.g. A(0,0), B(4,0), C(3,3), Q′(0.5, 0.5) gives PQ′ = BC/2 with Q′ not the midpoint.

To prove a tiling, we must show (a) every point of the plane is covered and (b) no two tiles overlap.

Method 1 (half-turns about midpoints of sides): Let the 4-gon have vertices V₀, V₁, V₂, V₃. A half-turn about the midpoint of one side followed by a half-turn about the midpoint of the next side is a slide (translation) by a diagonal: for example, about the midpoints of V₀V₁ and V₁V₂ it slides everything by $\overrightarrow{V_0{V}_2}$. So all the tiles are: the 4-gon and its half-turned copy, slid by every combination of whole multiples of the two diagonals. These pairs form a pattern that repeats like a parallelogram grid. Around each vertex the four different angles of the 4-gon appear once each, summing to 360°, so the tiles close up around every vertex without gaps or overlaps, and the repeating grid then covers the whole plane.

Method 2 (Varignon grid): the Varignon parallelograms tile the plane (sliding grid). Each 4-gon is made of its Varignon parallelogram plus four corner triangles, and each corner triangle is exactly the half-turn of a corner triangle of the neighbouring 4-gon about the shared side's midpoint. So the triangles fill the gaps between the parallelograms exactly, and the 4-gons cover the plane once.

Method 3 (slides along diagonals): this produces the same set of slid copies as Method 1 (slides by whole multiples of the two diagonals), and the gaps are the half-turned copies; so it gives the same tiling, and the argument for Method 1 applies.

In all three methods the tiles are the 4-gon and its half-turn, slid by whole multiples of the two diagonals; around each vertex the four angles (sum 360°) fit exactly, and the repeating grid covers the plane with no gaps or overlaps.

This is the same figure as Chapter 6, End-of-Chapter Q16 (Fig. 6.44). With the square from (0, 0) to (5, 5), the four lines enclose the square with vertices (1, 3), (3, 4), (4, 2), (2, 1), whose side is √5 and area 5, i.e. $\frac{5}{25}=\frac{1}{5}$ of the big square.

$\frac{1}{5}$ of the square is shaded.