Chapter 12Quadrilaterals

All three converses are true.

(a) Let AB = DC and AD = BC. Draw diagonal AC. Then △ACD ≅ △CAB (SSS: AD = CB, DC = BA, AC common). So ∠DCA = ∠BAC, which are alternate angles for AB and DC with transversal AC, giving AB ∥ DC. Similarly ∠DAC = ∠BCA gives AD ∥ BC. So ABCD is a parallelogram.

(b) Let ∠A = ∠C and ∠B = ∠D. Since ∠A + ∠B + ∠C + ∠D = 360°, we get 2(∠A + ∠B) = 360°, so ∠A + ∠B = 180°. These are co-interior angles for BC and AD with transversal AB, so AD ∥ BC. Similarly ∠B + ∠C = 180° gives AB ∥ DC.

(c) Let the diagonals meet at E with EA = EC and EB = ED. Then △AED ≅ △CEB (SAS, vertically opposite angles at E). So ∠DAE = ∠BCE — alternate angles — giving AD ∥ BC. Similarly △AEB ≅ △CED gives AB ∥ DC.

Compare △APQ and △CRQ:

∠AQP = ∠CQR (vertically opposite angles)

∠PAQ = ∠RCQ (alternate angles, since CR ∥ BA and AC is a transversal)

AP = CR (because CR = BP and BP = PA)

Two angles and a side not between them are equal, so △APQ ≅ △CRQ by AAS.

If P is any point on AB and PQ ∥ BC meets AC at Q, then Q divides AC in the same ratio in which P divides AB:

When P is the midpoint, AP/PB = 1, so AQ/QC = 1 and Q is the midpoint — exactly Theorem 7. This general result is the Basic Proportionality Theorem (Thales' Theorem). End-of-Chapter Exercise 11 proves it when AP/PB is rational.

From the Midpoint Theorem in △ABC and △MBC: PQ ∥ BC, XY ∥ BC, so PQ ∥ XY, and PQ = XY = BC/2.

P and Y both lie on median CP, and Q and X both lie on median BQ.

∠MPQ = ∠MYX (alternate angles: PQ ∥ XY, transversal PY)

PQ = YX (both equal BC/2)

∠MQP = ∠MXY (alternate angles: PQ ∥ XY, transversal QX)

The equal side lies between the two equal angles in each triangle, so △MPQ ≅ △MYX by ASA.

Key fact. Rotating by 180° about a point X and then by 180° about a point Y is the same as sliding (translating) by twice the distance from X to Y, in the direction X → Y. So in this tiling there are only two kinds of copy: copies with the same orientation as SOME (translations of it) and copies turned through 180°.

(1) A new copy in the second step touches two neighbours from the first step. Rotating either neighbour by 180° about the midpoint of the shared side gives a turned copy whose position is fixed by the side it shares. Both neighbours already fit around the same vertex with angles 1, 2, 3, 4, so the two rotations place the missing angle at the same spot with the same sides along the same lines — they produce the same quadrilateral.

(2) Around every vertex of the growing pattern, each of the four angles appears exactly once, and they add to 360°. So neighbouring copies meet along whole sides with no gap and no overlap.

Which possibility from Fig. 12.27? The one in which each copy is a 180° rotation of its neighbours about the midpoints of the shared sides, so that diagonally opposite copies (same colour) are translates of each other and the angles 1, 2, 3, 4 go round the point in order. Compare the three pictures with this description.

(i) Sides adjacent to AB: BC and DA (each shares an end point with AB). Side opposite to AB: CD.

(ii) Angles adjacent to ∠A: ∠B and ∠D. Angle opposite to ∠A: ∠C.

(iii) Two sides of a quadrilateral are opposite if they have no common end point (they are not adjacent).

Two internal angles are opposite if their vertices are not the two end points of a common side (the vertices are not adjacent).

Let the vertex be A, with adjacent vertices B and D and opposite vertex C. The two sides AB and AD make two angles at A: the ordinary angle ∠DAB (less than 180°) and the reflex angle 360° − ∠DAB.

Definition. The internal angle of ABCD at A is

• the reflex angle 360° − ∠DAB, if A lies inside the triangle BCD formed by the other three vertices;

• the ordinary angle ∠DAB (less than 180°) otherwise.

Check with DART. The dent D lies inside △ART, so the internal angle at D is reflex (more than 180°). No other vertex of DART lies inside the triangle of the remaining three, so the other three internal angles are ordinary angles — as in Fig. 12.4.

For a convex quadrilateral, no vertex lies inside the triangle of the other three, so every internal angle is the ordinary angle, less than 180°.

AB ∥ DC — No, ABCD must be convex. A and B lie on one line, C and D on a parallel line. A non-convex quadrilateral needs a vertex (the dent) lying inside the triangle formed by the other three vertices. But any triangle formed from three of these vertices has two vertices on one of the parallel lines and one on the other, so its inside lies strictly between the two lines. The remaining vertex lies on one of the lines, so it cannot be inside. Hence there is no dent.

AB = CD — Yes, it can be non-convex. Take A(0, 0), B(5, 0), C(3, 2), D(0, 6). Then AB = 5 and \(CD = \sqrt{3^2 + 4^2} = 5\), but C lies inside △ABD, so the angle at C is reflex.

∠A = ∠C — Yes, it can be non-convex. Take an arrowhead (dart) that is symmetric about BD: A(−2, 0), B(0, 1), C(2, 0), D(0, 4). By symmetry ∠A = ∠C, while B lies inside △ACD, so the angle at B is reflex. (Only one angle of a quadrilateral can be reflex, since they add to 360°, so the dent must be at B or D, never at A or C.)

The three lines make 7 regions: the inside of △ABC, three regions across the sides (one beyond each side), and three regions at the vertices (inside the vertically opposite angle at A, B or C). Remember the sides of ABCD are AB, BC, CD, DA.

D inside △ABC: non-convex (dent at D).

D across side AC (opposite side of AC from B): convex.

D across side AB: segment CD crosses segment AB — self-intersecting.

D across side BC: segment AD crosses segment BC — self-intersecting.

D in the vertex region at A: A lies inside △BCD — non-convex (dent at A).

D in the vertex region at C: C lies inside △ABD — non-convex (dent at C).

D in the vertex region at B: B lies inside △ACD — non-convex (dent at B).

No. In a self-intersecting 4-gon, two opposite sides, say AB and CD, cross at a point E. Two lines that meet at a point always lie in one plane. That plane contains line AB and line CD, and hence all four vertices A, B, C, D. So the 4-gon is planar.

(i) True. Adjacent angles of a parallelogram add up to 180°. If one angle is 90°, its neighbours are 90°, and the opposite angle equals it, so all four angles are 90°.

(ii) False. The diagonals of every rhombus are perpendicular, so this condition adds nothing. Counterexample: a rhombus with angles 60° and 120° has perpendicular diagonals but is not a square.

(iii) True. In parallelogram ABCD with AC = BD: △ABC ≅ △DCB (SSS: AB = DC, BC common, AC = DB). So ∠B = ∠C. But ∠B + ∠C = 180° (adjacent angles), so ∠B = ∠C = 90° and ABCD is a rectangle.

Given: ABCD is a parallelogram and ∠BAC = ∠DAC.

AB ∥ DC with transversal AC: ∠BAC = ∠DCA (alternate angles).

AD ∥ BC with transversal AC: ∠DAC = ∠BCA (alternate angles).

Since ∠BAC = ∠DAC, we get ∠DCA = ∠BCA, so AC bisects ∠C.

Also ∠DAC = ∠BAC = ∠DCA. In △ADC, the angles at A and C are equal, so DA = DC (sides opposite equal angles).

A parallelogram with two adjacent sides equal has all four sides equal (opposite sides are equal), so ABCD is a rhombus.

(i) Yes. AC bisects ∠A and ∠C, so △ABC ≅ △ADC (ASA: ∠BAC = ∠DAC, AC common, ∠BCA = ∠DCA). Hence AB = AD and CB = CD. In the same way BD bisects ∠B and ∠D, so △BAD ≅ △BCD, giving BA = BC and DA = DC. So AB = BC = CD = DA — a rhombus.

(ii) Yes. Diagonals that bisect each other make ABCD a parallelogram (Theorem 4). Let them meet at O. △AOB ≅ △AOD (SAS: OB = OD, ∠AOB = ∠AOD = 90°, OA common), so AB = AD. A parallelogram with equal adjacent sides is a rhombus.

(iii) No. Counterexample: an isosceles trapezium (non-parallel sides equal) has equal diagonals but is not a rectangle.

Extra condition: the diagonals are equal and bisect each other. Then ABCD is a parallelogram with equal diagonals, which is a rectangle (Q1(iii)).

Consider the bisectors of two adjacent angles, say ∠A and ∠D, meeting at a point (a vertex of the new quadrilateral). Since AB ∥ DC, ∠A + ∠D = 180°, so

In the triangle formed by AD and these two bisectors, the third angle is 180° − 90° = 90°.

The same holds for the bisectors of ∠A and ∠B, of ∠B and ∠C, and of ∠C and ∠D. So each of the four intersection points is a vertex with a 90° angle (either this angle itself or its vertically opposite angle). A quadrilateral with all four angles 90° has its opposite angles equal, so it is a parallelogram with a right angle — a rectangle.

Why AB ≠ BC? If AB = BC, ABCD is a rhombus. Then the angle bisectors are the diagonals, and all four bisectors pass through a single point (the centre). The four intersection points coincide, so there is no rectangle.

(i) By the Midpoint Theorem:

PQ = BC/2 = BR = RC,   QR = AB/2 = AP = PB,   RP = AC/2 = AQ = QC.

△PQR and △QPA: PQ = QP, QR = PA, RP = AQ, so △PQR ≅ △QPA (SSS).

△PQR and △RBP: PQ = RB, QR = BP, RP = PR, so △PQR ≅ △RBP (SSS).

△PQR and △CRQ: PQ = CR, QR = RQ, RP = QC, so △PQR ≅ △CRQ (SSS).

(ii) Yes. Through each vertex of △PQR, draw the line parallel to the opposite side. The three lines meet in pairs at A, B, C.

Reason: QR ∥ AB and P lies on AB, so AB is the line through P parallel to QR. Similarly AC is the line through Q parallel to PR, and BC is the line through R parallel to PQ. Since there is only one line through a point parallel to a given line, these three lines are exactly the sides of the erased triangle.

Let MN meet AD at X.

By the Midpoint Theorem in △ABC, MN ∥ BC. Since D lies on BC, MX ∥ BD.

In △ABD, M is the midpoint of AB and the line MX through M is parallel to BD. By Theorem 7 (converse of the Midpoint Theorem), X is the midpoint of AD.

Let M be the midpoint of AD.

In △ADC: M is the midpoint of AD, G is the midpoint of AC, so MG ∥ DC (Midpoint Theorem).

In △DAB: M is the midpoint of DA, H is the midpoint of DB, so MH ∥ AB ∥ DC.

So MG and MH are both lines through M parallel to DC. There is only one such line, so M, G, H are collinear and the line GH is parallel to DC, and hence to AB.

Why AB ≠ CD? If AB = CD as well as AB ∥ DC, then ABCD is a parallelogram (Theorem 5). Its diagonals bisect each other, so G = H and there is no segment GH.

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

(ii) By the Midpoint Theorem in △ABC and △BCD: PQ = AC/2 and QR = BD/2. 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 — true. If PR ⊥ QS, the parallelogram PQRS has diagonals that bisect each other at right angles, so it is a rhombus (Ex 12.2, Q3(ii)). Then PQ = QR, i.e. AC/2 = BD/2, so AC = BD.

(i) Construction.

Place A′: construct △A′PS congruent to △APS on the new paper, on the same side of PS as A (copy the lengths AP and AS from the original).

Place B′: extend A′P beyond P to B′ with PB′ = A′P (P is to be the midpoint of A′B′).

Place C′: extend B′Q beyond Q to C′ with QC′ = B′Q.

Place D′: extend C′R beyond R to D′ with RD′ = C′R.

(ii) Why it works.

△A′PS ≅ △APS (by construction), so ∠A′PS = ∠APS. Since A′, P, B′ are collinear (and A, P, B are too), ∠B′PQ = 180° − ∠A′PS − ∠SPQ = ∠BPQ. Also PB′ = PA′ = PA = PB. So △B′PQ ≅ △BPQ (SAS), giving QB′ = QB and ∠PQB′ = ∠PQB.

Repeating the same argument at Q and then at R gives △C′QR ≅ △CQR and then △D′RS ≅ △DRS.

S is collinear with A′ and D′: ∠A′SP + ∠PSR + ∠RSD′ = ∠ASP + ∠PSR + ∠RSD = 180°, because A, S, D are collinear. So A′, S, D′ are collinear, and SA′ = SA = SD = SD′, i.e. S is the midpoint of A′D′.

Congruence: A′B′ = 2PA = AB, B′C′ = 2QB = BC, C′D′ = 2RC = CD, D′A′ = 2SD = DA. The angles match too: ∠D′A′B′ = ∠SA′P = ∠SAP = ∠DAB, and similarly at B′, C′, D′ from the congruent triangles. All sides and all angles are equal, so A′B′C′D′ ≅ ABCD.

(iii) PQ ∥ AC and PQ = AC/2 (△ABC); QR ∥ BD and QR = BD/2 (△BCD).

If PQRS is a square: PQ = QR gives AC = BD, and PQ ⊥ QR gives AC ⊥ BD.

Converse: if AC = BD and AC ⊥ BD, then PQ = QR and PQ ⊥ QR. A parallelogram with equal adjacent sides and a right angle is a square.

Each interior angle of a regular pentagon is \(\dfrac{(5-2)\times 180^\circ}{5} = 108^\circ\).

Take any vertex of a pentagon in the tiling. The angles around that point must total 360°.

Case 1: the point is a vertex of every tile around it. Then 360° must be a sum of 108° angles. But 3 × 108° = 324° (a gap of 36°) and 4 × 108° = 432° (overlap). 360 is not a multiple of 108.

Case 2: the point lies on the side of one tile (a straight angle, 180°). Then the remaining 180° must be made of 108° angles — but one gives 108° and two give 216°. Impossible.

So the angles around a vertex can never add to exactly 360°, and no tiling exists.

The four angles of DART still add up to 360° (one of them, at D, is reflex), so the same ideas work.

Method 1. Take a copy of DART and rotate a second copy by 180° about the midpoint of one side, so the two copies share that side. Repeat at every free side of every copy. Around each vertex the angles ∠D, ∠A, ∠R, ∠T appear once each (the reflex angle at D together with the three small angles make 360°), so there are no gaps or overlaps. The dents of one copy are filled by the "points" of its neighbours.

Method 2. Draw the Varignon parallelogram of DART (joining the midpoints of its sides — it is a parallelogram even for a non-convex 4-gon) and make a grid of copies of it. Place copies of DART on alternate parallelograms, matching the midpoints of their sides with the grid points; the gaps are automatically filled by turned copies of DART.

Which is better? Method 1 is usually easier to carry out with paper cutouts, since each step uses only a half-turn about a side's midpoint. Method 2 is nice because the grid is fixed in advance, so the whole pattern can be drawn directly. Either choice is acceptable with a reason.

Take △ABC and a second copy. Rotate the second copy through 180° about the midpoint M of side BC, so B and C swap places and A goes to a point A′ with M the midpoint of AA′.

In quadrilateral ABA′C, the diagonals AA′ and BC bisect each other at M, so it is a parallelogram (Theorem 4).

Copies of any parallelogram tile the plane (the grid of parallel sticks in the introduction). Each parallelogram in that tiling is made of two copies of △ABC, split by a diagonal. So the triangle tiles the plane.

The drawn segment XY starts on the 2nd line and ends on the 6th line — it crosses 4 gaps. Its midpoint is M, the point where it crosses the 4th line (the middle line).

Justification. Through M, draw the line perpendicular to the ruled lines, meeting line 2 at U and line 6 at V. Since the lines are equally spaced, MU = 2 gaps = MV.

In △MUX and △MVY:

MU = MV

∠MUX = ∠MVY = 90°

∠XMU = ∠YMV (vertically opposite angles)

So △MUX ≅ △MVY (ASA), and MX = MY. Hence M is the midpoint of XY.

AB and CD cross at E, so the figure splits into two triangles, △EAD and △EBC, with ∠AED = ∠BEC (vertically opposite).

Since E lies on AB and on CD: ∠A = ∠DAE, ∠D = ∠ADE, ∠B = ∠EBC, ∠C = ∠ECB.

In △EAD: ∠A + ∠D = 180° − ∠AED.

In △EBC: ∠B + ∠C = 180° − ∠BEC = 180° − ∠AED.

Adding: ∠A + ∠B + ∠C + ∠D = 360° − 2∠AED < 360°, since ∠AED > 0°.

Sum = 2°: we need 360° − 2∠AED = 2°, i.e. ∠AED = 179°. Draw two lines through E meeting at an angle of 1° (so the other pair of vertically opposite angles is 179°). Take A and B on the first line on opposite sides of E, and D and C on the second line on opposite sides of E, with A and D in the 179° angle. Then ∠AED = ∠BEC = 179°, and the angle sum is exactly 2°.

Let the diagonals AC and BD meet at O. Since ABCD is a parallelogram, OA = OC and OB = OD.

OP = OD − DP and OQ = OB − BQ. Since OD = OB and DP = BQ, OP = OQ.

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

In Fig. 12.35 the right angle is at B, so AB ⊥ BC.

When A is folded onto B, every point of the crease is equidistant from A and B, so the crease is the perpendicular bisector of AB. It passes through the midpoint M of AB and is perpendicular to AB.

BC is also perpendicular to AB. Two lines perpendicular to the same line are parallel, so the crease is parallel to BC.

So the crease is the line through the midpoint M of AB, parallel to BC. By Theorem 7 it meets AC at the midpoint of AC.

(i) P is the midpoint of BC. Turn △MPB through 180° about P. Then B goes to C (since PB = PC) and M goes to a point M′ with P the midpoint of MM′.

It is a triangle: M, P, M′ are collinear by construction (∠MPC + ∠CPM′ = ∠MPC + ∠MPB = 180°, as B, P, C are collinear). So △MPC and the moved piece △M′PC together form △MM′C.

Side lengths: by the Centroid Theorem, MP = AP/3 and AM = 2AP/3, so

MM′ = 2MP = AM = \(\tfrac{2}{3}AP\),   M′C = MB = \(\tfrac{2}{3}BQ\),   MC = \(\tfrac{2}{3}CR\).

Pairing the other pieces in the same way (△MQC with △MQA about Q, and △MRA with △MRB about R) gives two more triangles with sides AM, BM, CM. All three are congruent by SSS: each has sides equal to \(\tfrac{2}{3}\) of the three medians.

(ii) Look at the medians of the assembled △MM′C. MBM′C is a parallelogram (its diagonals MM′ and BC bisect each other at P).

• Median from C goes to P, the midpoint of MM′: CP = BC/2.

• Median from M goes to N, the midpoint of M′C. Here NC = M′C/2 = MB/2 = MQ, and NC ∥ MQ, pointing the same way (along BQ). So MQCN is a parallelogram and MN = QC = AC/2.

• In the same way the median from M′ equals AB/2.

So the medians of each assembled triangle are BC/2, AC/2, AB/2. Applying (i) to it produces triangles with sides \(\tfrac{2}{3}\) of these medians:

This holds for each of the 3 assembled triangles, giving 9 triangles with sides AB/3, BC/3, AC/3.

(iii) Cut along median AP to get △APC and △APB. In △APB, draw the median from P to K, the midpoint of AB, and cut along PK. Turn △PKB through 180° about K: B goes to A and P goes to P′, with K the midpoint of PP′.

△PKA and the moved piece form △APP′ (P, K, P′ collinear). Its sides are AP, AP′ = BP = PC, and PP′ = 2PK = AC (Midpoint Theorem in △ABC: KP = AC/2).

So △APP′ has sides AP, PC, AC — the same as △APC. The two triangles are congruent (SSS).

(i) Draw diagonal AC, meeting EF at G.

In △ADC: E is the midpoint of AD and EG ∥ DC, so G is the midpoint of AC and EG = DC/2 (Theorem 7).

In △CAB: G is the midpoint of CA and GF ∥ AB, so F is the midpoint of CB and GF = AB/2 (Theorem 7).

So F is the midpoint of BC, and

(ii) Way 1 (direct). Let M be the midpoint of BD. In △DAB, E and M are midpoints of DA and DB, so EM ∥ AB. In △BDC, M and F are midpoints of BD and BC, so MF ∥ DC ∥ AB. EM and MF are both lines through M parallel to AB, so they are the same line. Hence E, M, F are collinear and EF ∥ AB.

Way 2 (using (i)). Draw the line through E parallel to AB; it meets BC at some point F′. By (i), F′ is the midpoint of BC, so F′ = F. So EF is that parallel line: EF ∥ AB.

Way 2 is shorter, since (i) does all the work — but either is fine.

Compare △OAP and △OCQ:

OA = OC (diagonals of a parallelogram bisect each other)

∠OAP = ∠OCQ (alternate angles: AB ∥ DC, transversal AC)

∠AOP = ∠COQ (vertically opposite angles)

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

This congruence proof is probably the simplest; another approach uses the half-turn about O, which maps the parallelogram onto itself and line AB onto line CD.

By Exercise 7(i), EF ∥ AD ∥ BC and

EF halves the height. Let h be the distance between AD and BC. Drop the perpendicular AK from A to BC; it meets EF at L. In △ABK, E is the midpoint of AB and EL ∥ BK, so L is the midpoint of AK. So each of the two trapeziums has height h/2.

Area of AEFD = \(\frac{1}{2}(3 + 4)\cdot\frac{h}{2} = \frac{7h}{4}\)

Area of EBCF = \(\frac{1}{2}(4 + 5)\cdot\frac{h}{2} = \frac{9h}{4}\)

(i) 3 quadrilaterals. A quadrilateral is fixed by the cyclic order of its vertices. Starting from A, the other three can be arranged in 3! = 6 orders, but each quadrilateral is counted twice (once in each direction). So there are 6 ÷ 2 = 3: ABCD, ABDC and ACBD.

(ii) Two cases.

Case 1: no point lies inside the triangle formed by the other three. Exactly 1 quadrilateral is convex (going round the outside), and the other 2 are self-intersecting (their "diagonals" become sides that cross).

Case 2: one point lies inside the triangle formed by the other three. All 3 quadrilaterals are non-convex (each has its dent at the inside point); none is convex and none is self-intersecting.

Key fact (equal intercepts). If points on AB are equally spaced and lines through them are drawn parallel to BC, they cut AC into equal parts. For the first step this is Theorem 7 (in a triangle); for each later step it is Exercise 7(i) (in a trapezium).

(i) Mark X on PB with PX = XB = 1. Then A, P, X, B are equally spaced (gaps of 1). The parallels to BC through P and X meet AC at Q and Y.

In △AXY: P is the midpoint of AX and PQ ∥ XY, so AQ = QY = \(\sqrt5\).

In trapezium PQCB: X is the midpoint of PB and XY ∥ BC, so Y is the midpoint of QC: QY = YC = \(\sqrt5\).

So QC = \(2\sqrt5\).

(ii) Divide AP into 3 equal parts of length \(\frac13\). Then AB is made of 4 equal parts of \(\frac13\): 3 in AP and 1 in PB. The parallels cut AC into 4 equal parts, 3 of which make AQ = \(\sqrt5\). Each part is \(\frac{\sqrt5}{3}\), so QC = \(\frac{\sqrt5}{3}\).

(iii) Again use parts of length \(\frac13\): AP has 3 parts and PB = \(\frac23\) has 2 parts. AQ has 3 equal parts, each \(\frac{\sqrt5}{3}\), and QC has 2 of them: QC = \(\frac{2\sqrt5}{3}\).

(iv) Suppose \(\frac{AP}{PB} = \frac{m}{n}\) with m, n positive whole numbers. Let \(d = \frac{AP}{m} = \frac{PB}{n}\). Divide AP into m parts and PB into n parts, each of length d. Through all the division points, draw lines parallel to BC. By the equal-intercept fact, AC is cut into m + n equal parts; m of them make AQ and n make QC. So

(i) MB = AB/2 = DC/2 = DN, and MB ∥ DN. So MBND is a parallelogram (Theorem 5), which gives DM ∥ BN.

Let DM meet AC at X, and BN meet AC at Y.

In △ABY: M is the midpoint of AB and MX ∥ BY, so X is the midpoint of AY. Thus AX = XY.

In △CDX: N is the midpoint of CD and NY ∥ DX, so Y is the midpoint of CX. Thus CY = YX.

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

(ii) Way 1 (using (i)). Make PQ a diagonal of a parallelogram: find the midpoint O of PQ, draw any line through O and mark B and D on it with OB = OD. Then PBQD is a parallelogram. Join D to the midpoint of PB and B to the midpoint of QD; these cut PQ at its trisection points.

Way 2 (Exercise 11). Draw a ray from P and mark three equal lengths PX₁ = X₁X₂ = X₂X₃ with a compass. Join X₃Q. Draw lines through X₁ and X₂ parallel to X₃Q. They cut PQ into three equal parts (equal intercepts).

Way 3 (Centroid Theorem). Make PQ a median: draw any line through Q and mark B and C on it with QB = QC. In △PBC, PQ is a median. Draw a second median, from B to the midpoint of PC; it meets PQ at the centroid G, with PG = 2GQ. So GQ = PQ/3. The midpoint of PG gives the other trisection point.

(i) The proof never used convexity. It only applied the Midpoint Theorem to the four triangles ABC, ADC, BCD and BAD. These are genuine triangles whatever the shape of ABCD (no three vertices are collinear), so PQ ∥ AC ∥ SR and QR ∥ BD ∥ PS still hold, with PQ = SR = AC/2. Hence PQRS is a parallelogram for non-convex and self-intersecting ABCD too.

(ii) PQ ∥ AC and QR ∥ BD. If the lines AC and BD are parallel, then PQ and QR lie along the same line, and P, Q, R, S are all collinear — the "parallelogram" collapses into a segment.

This can only happen for a self-intersecting ABCD (in a convex quadrilateral the diagonals cross, and in a non-convex one the line through the dent vertex crosses the other diagonal). Example: A(0, 0), B(0, 2), C(4, 0), D(4, 2). Here AC and BD are both horizontal. The midpoints are P(0, 1), Q(2, 1), R(4, 1), S(2, 1) — all on the line y = 1.

(iii) Yes. Each of the four triangles ABC, ADC, BCD, BAD still lies in its own plane, so the Midpoint Theorem applies to each. PQ and SR are both parallel to AC and equal to AC/2. Two lines parallel to the same line are parallel to each other (in space too), so PQ ∥ SR with PQ = SR. Parallel lines lie in one plane, so PQRS is a planar parallelogram — even though ABCD is not planar.

No. Counterexample: let ABCD be a square of side 4. Take P on AB, Q on BC, R on CD, S on DA with AP = BQ = CR = DS = 1.

Then each of the four corner triangles (e.g. △SAP) is right-angled with legs 1 and 3, so they are all congruent (SAS). Hence PQ = QR = RS = SP, and PQRS is a rhombus — in fact a square — so it is certainly a parallelogram.

But AP = 1 ≠ 2 = AB/2, so P is not the midpoint of AB (and similarly for Q, R, S).

(i) Extend PQ to S so that QS = PQ (Q becomes the midpoint of PS).

Why we want △APQ ≅ △CSQ: it would give CS = AP (= PB) and ∠QCS = ∠QAP, i.e. CS ∥ AB. That makes PBCS a parallelogram, which is exactly what yields PQ ∥ BC and PQ = BC/2.

Proof. In △APQ and △CSQ: AQ = CQ (Q is the midpoint of AC), PQ = SQ (construction), ∠AQP = ∠CQS (vertically opposite). So △APQ ≅ △CSQ (SAS).

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

PBCS has one pair of opposite sides (PB, CS) equal and parallel, so it is a parallelogram (Theorem 5). Therefore PS ∥ BC and PS = BC. Since PQ = PS/2, we get PQ ∥ BC and PQ = BC/2.

(ii) Converse. Now P is the midpoint of AB and PQ ∥ BC, with Q on AC. Extend PQ to S so that PS = BC.

PS ∥ BC and PS = BC, so PBCS is a parallelogram (Theorem 5). Hence CS ∥ PB (i.e. CS ∥ AB) and CS = PB = AP.

In △APQ and △CSQ: AP = CS, ∠PAQ = ∠SCQ (alternate angles, AB ∥ CS, transversal AC), ∠APQ = ∠CSQ (alternate angles, transversal PS). So △APQ ≅ △CSQ (ASA).

Therefore AQ = CQ — Q is the midpoint of AC — and PQ = QS = PS/2 = BC/2.

Here are several properties with their converses (for a general quadrilateral ABCD).

Rhombus

(a) If a quadrilateral is a rhombus, its diagonals bisect each other at right angles. Converse: if the diagonals bisect each other at right angles, it is a rhombus. True (Ex 12.2 Q3(ii)).

(b) If it is a rhombus, its diagonals bisect its angles. Converse: if the diagonals bisect the angles, it is a rhombus. True (Ex 12.2 Q3(i)).

(c) If it is a rhombus, its diagonals are perpendicular. Converse: if the diagonals are perpendicular, it is a rhombus. False — a kite has perpendicular diagonals.

Rectangle

(d) If it is a rectangle, its diagonals are equal. Converse: if the diagonals are equal, it is a rectangle. False — an isosceles trapezium. (True if it is already known to be a parallelogram.)

(e) If it is a rectangle, its diagonals are equal and bisect each other. Converse: True (parallelogram with equal diagonals).

Square

(f) If it is a square, its diagonals are equal and perpendicular. Converse: False — the kite A(0, 2), B(−1, 0), C(0, −2), D(3, 0) has diagonals AC = BD = 4, perpendicular, but it is not a square.

(g) If it is a square, its diagonals are equal, perpendicular and bisect each other. Converse: True — bisecting diagonals make it a parallelogram, equal diagonals a rectangle, and perpendicular diagonals a rhombus; a rectangle that is a rhombus is a square.

Start with a flat convex quadrilateral ABCD (angle sum 360°) and treat the diagonal AC as a hinge. Keep △ADC fixed and rotate △ABC about AC, lifting B out of the plane.

∠B and ∠D do not change — each belongs to one rigid triangle.

∠A and ∠C decrease. The angle ∠DAB sits in △DAB, whose sides AD and AB stay the same length. As B turns about AC, it moves closer to D (in the flat convex position B and D are as far apart as possible, on opposite sides of AC). With two sides fixed, a shorter third side BD means a smaller angle opposite it, so ∠DAB decreases. In the same way ∠BCD (in △BCD with fixed CB, CD) decreases.

So as soon as ABCD becomes non-planar, ∠A + ∠B + ∠C + ∠D < 360°. Every non-planar quadrilateral can be obtained this way (fold it along a diagonal back into the plane), so its angle sum is always less than 360°.

Sum = 2°: Yes. Choose two long, thin triangles on AC — for example with ∠B = ∠D = 0.4°, ∠BAC = 90°, ∠DAC = 89.7°, ∠BCA = 89.6°, ∠DCA = 89.9°. As B is rotated all the way over to D's side, the angle sum falls continuously from 360° down to 0.4° + 0.4° + (90° − 89.7°) + (89.9° − 89.6°) = 1.4°. So somewhere during the rotation, while ABCD is non-planar, the sum is exactly 2°.

(i) Sliding the cutout along EO′ by the length EO moves every point of it the same distance EO in the same direction. So the slid copy has vertices E′, O′ and two points S′, M′ with SS′ ∥ EO, SS′ = EO and MM′ ∥ EO, MM′ = EO (the dashed arrows in Fig. 12.42). Since EO and E′O′ lie on one line with E′ = O, also EE′ = OO′ = EO.

Now look at four quadrilaterals, each with one pair of opposite sides equal and parallel to EO:

ESS′E′ (ES and E′S′), OSS′O′ (OS and O′S′), EMM′E′ (EM and E′M′), OMM′O′ (OM and O′M′).

By Theorem 5 each is a parallelogram, so their other pairs of opposite sides are equal and parallel:

E′S′ = ES,   O′S′ = OS,   E′M′ = EM,   O′M′ = OM.

Each side of S′O′M′E′ is equal and parallel to the matching side of SOME, so S′O′M′E′ is an exact shifted copy of SOME — the cutout matches exactly.

(ii) Sliding along diagonal EO (both directions) and along diagonal SM (both directions) gives 4 new copies around the starting copy, each touching it only at a vertex. Repeating from these 4 fills in the corners of a 3 × 3 grid of 9 copies, all shifted copies of SOME. The gaps between them are also 4-gons congruent to SOME (turned through 180°), so together the two kinds of copies cover the plane.

(i) Draw BC = c. Draw a circle of radius b about B and a circle of radius a about C (a ≤ b ≤ c). The circles meet (giving the third vertex A, off the line BC) exactly when a + b > c. If a + b = c they only touch on BC (no triangle), and if a + b < c they do not meet at all.

(ii) A 4-gon exists exactly when the longest side is less than the sum of the other three (see (iii)). Let the fourth side be x.

x = 100: is 100 < 2 + 5 + 11 = 18? No — not possible.

x = 10: longest is 11, and 11 < 2 + 5 + 10 = 17 — possible.

x = 1: longest is 11, and 11 < 2 + 5 + 1 = 8? No — not possible.

In general we need 11 < 7 + x (if x ≤ 11) and x < 18 (if x ≥ 11), i.e. 4 < x < 18.

(iii) Rule: a 4-gon with sides a, b, c, d exists exactly when the largest of them is less than the sum of the other three.

Necessary: say d is the largest, with sides in order a, b, c, d, and draw the diagonal e between the ends of a, b. Triangle inequalities give d < c + e and e < a + b, so d < a + b + c.

Sufficient: if d < a + b + c, choose a diagonal length e with both d − c < e < a + b and |a − b| < e < c + d (possible since d − c < a + b and |a − b| < c + d). Then triangles with sides (a, b, e) and (c, d, e) both exist; build them on opposite sides of a segment of length e to get the 4-gon.

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

Pattern: the differences are 2, 3, 4, 5, 6 — each time one more.

(ii) Formula:

Going from n to n + 1 sides adds n − 1 diagonals: the new vertex is joined to the n − 2 old vertices that are not its neighbours, and the old side joining its two neighbours becomes a diagonal. Check: \(\frac{(n+1)(n-2)}{2} - \frac{n(n-3)}{2} = \frac{2n-2}{2} = n - 1\).

Justification for all n: each vertex is joined by a diagonal to every vertex except itself and its two neighbours, i.e. to n − 3 vertices. That gives n(n − 3) ends of diagonals, and each diagonal has 2 ends, so there are n(n − 3)/2 diagonals.

Formula: sum of the angles = (n − 2) × 180°.

Proof (convex n-gon). Join one vertex to all the other vertices. The n − 3 diagonals from that vertex split the polygon into n − 2 triangles. The angles of these triangles together make up exactly the angles of the polygon, so the sum is (n − 2) × 180°.

Any (non-convex) n-gon. Every such polygon with n ≥ 4 has a diagonal lying completely inside it. That diagonal splits it into a k-gon and an (n − k + 2)-gon, both with fewer sides. If the formula holds for those, the angle sum is (k − 2) × 180° + (n − k) × 180° = (n − 2) × 180°. Starting from triangles, this proves the formula for all n.

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

Proof. Draw the line through C parallel to BA, meeting line PQ at R. PR ∥ BC and CR ∥ BP, so PBCR is a parallelogram and PR = BC = 2PQ. So Q is the midpoint of PR: PQ = QR.

In △APQ and △CRQ: ∠AQP = ∠CQR (vertically opposite), ∠APQ = ∠CRQ (alternate angles, AB ∥ CR), PQ = RQ. So △APQ ≅ △CRQ (AAS).

Hence AQ = CQ, so Q is the midpoint of AC; and AP = CR = PB (opposite sides of parallelogram PBCR), so P is the midpoint of AB.

(ii) Converse of the second sentence: “Suppose P is the midpoint of side AB of △ABC and Q is a point on side AC. If PQ ∥ BC then Q is the midpoint of AC.” This is exactly Theorem 7. In named conditions: If (P MID) and (PRLL), then (Q MID) and (HALF).

(iii) There are 6 ways to choose two of the four conditions:

{P MID, Q MID} — the Midpoint Theorem; {PRLL, HALF} — part (i); {P MID, PRLL} — Theorem 7; {Q MID, PRLL} — Theorem 7 with B and C swapped; {Q MID, HALF} — same as {P MID, HALF} with B and C swapped. So only {P MID, HALF} remains.

In words: “If P is the midpoint of AB and Q is a point on AC with PQ = BC/2, must Q be the midpoint of AC, and must PQ be parallel to BC?”

Answer: not necessarily. The midpoint N of AC always has PN = BC/2, but a circle of radius BC/2 about P can meet segment AC at a second point. Example: A(0, 0), B(3, 2), C(4, 0). Then P = (1.5, 1), BC = \(\sqrt5\), and Q = (1, 0) lies on AC with \(PQ = \sqrt{0.25 + 1} = \frac{\sqrt5}{2} = \frac{BC}{2}\). But Q is not the midpoint (2, 0) of AC, and PQ is not parallel to BC. So neither conclusion follows.

What must be shown: (a) every point of the plane lies in some copy (no gaps), and (b) no two copies share an inside point (no overlaps).

One picture behind all three methods. Let SOME be the 4-gon and T its 180° rotation about the midpoint of side OM. Together, SOME and T form a hexagon-like piece H whose opposite sides are equal and parallel. Translating H by the two diagonals EO and SM of SOME (and their combinations) moves it to positions that fit together along whole sides.

Method 1 (half-turns). Two half-turns make a translation, so every copy produced is either SOME shifted by a combination of the diagonal vectors EO and SM, or T shifted by such a combination. These are exactly the copies of H arranged on a parallelogram grid built from EO and SM. Around every vertex the four angles 1, 2, 3, 4 appear once each and add to 360°, so neighbouring copies meet along whole sides without gaps or overlaps. The grid of translations covers the whole plane, so the copies do too.

Method 2 (Varignon grid). The sides of the Varignon parallelogram are half the diagonals of SOME. The copies placed on alternate parallelograms are SOME shifted by combinations of EO and SM, and Exercise Set 12.3 Q5 shows that each gap is surrounded by four side-midpoints that determine a copy of SOME exactly (turned through 180°). So this gives the same set of copies as Method 1.

Method 3 (sliding along diagonals). Exercise 18(i) shows each slide is the translation by EO or SM. So Method 3 also produces SOME shifted by combinations of EO and SM, and the blank spaces are the turned copies — the same tiling again.

Since all three procedures build the same arrangement, and that arrangement has no gaps or overlaps at any vertex and repeats along a parallelogram grid covering the plane, each method gives a tiling.

Let the square have side s. Each of the four lines joins a vertex to the midpoint of a side, as marked in Fig. 12.43. The lines form two pairs of parallel lines, and the two pairs are perpendicular (each pair is the other pair turned through 90° about the centre of the square), so the shaded figure is a square.

Distance between two parallel lines. Take the pair that starts at the bottom-left corner and at the midpoint of the left side. Between them, inside the big square, is a parallelogram with two vertical sides of length s/2 (on the left and right sides of the square), a horizontal width of s, and slanted sides from a vertex to a side midpoint.

Its area = (vertical side) × (horizontal distance) = \(\frac{s}{2}\times s = \frac{s^2}{2}\).

The slanted side has length \(\sqrt{s^2 + \left(\frac{s}{2}\right)^2} = \frac{\sqrt5}{2}s\). So the distance d between the two parallel lines satisfies

The shaded square's side is this distance, so its area is \(\left(\frac{s}{\sqrt5}\right)^2 = \frac{s^2}{5}\).

Check by rearranging: the four small right triangles outside the shaded square can each be moved to complete a copy of the shaded square, so the big square is made up of exactly 5 shaded-size squares.