Quadrilaterals and parallelograms: Questions and Answers

78 quadrilaterals and parallelograms questions solved step by step. Open a question for the full working, the marks for each step and exam practice.

• Can we use any given quadrilateral to tile the plane? If not, which quadrilaterals can be used and which cannot?Answer: Yes. Copies of any quadrilateral, regular or irregular, convex or with a dent, can tile the plane. Turn a copy through 180° about the midpoint of a side and it fits exactly along that side; repeating this fills the plane, and at every corner the four different angles meet, adding up to 360°.
• Informally, a quadrilateral is a figure with four straight sides, as in the first figure ABCD in the figure below. But consider the other six figures in the figure: the five plane figures NOPE, SILY, DART, CUTS, OPENS and the non-planar BENT. Should we call all these figures quadrilaterals? If you answer ‘no’ for any of them, how will you define a quadrilateral so that such a figure is excluded? As you can see, some care is needed to precisely define what we think of as a quadrilateral.Answer: No. Besides ABCD, only DART is accepted (as a non-convex quadrilateral). NOPE (three vertices in a line), SILY (overlapping sides), CUTS (sides cross), OPENS (not closed, five points) and BENT (not in one plane) are excluded by the definition: four distinct points A, B, C, D in a plane, no three collinear, joined by AB, BC, CD, DA, with every point other than A, B, C, D lying on exactly one of these four segments.
• To prepare, let us first consider how we can define a triangle. Let A, B and C be three points. Can we say that ∆ABC consists of points on the three line segments AB, BC, CA?Answer: Yes, provided A, B, C are not collinear. If the three points lie on one line, the three segments just make one line segment, not a triangle.
• Can we similarly define a quadrilateral ABCD?Answer: Not with just one condition. For a triangle, “not collinear” is enough. For ABCD we need three conditions: no three of A, B, C, D collinear; all four in one plane; and the sides AB, BC, CD, DA meet only at their shared ends (every non-vertex point is on exactly one side).
• Take any 4 distinct non-collinear points A, B, C and D. Will segments AB, BC, CD and DA always form a quadrilateral as we visualise it?Answer: No. Even when the four points are not all on one line, three of them may be collinear (giving a triangle or overlapping sides), the four may not lie in one plane, or two sides may cross.
• Can you come up with a definition that rules out self-intersecting quadrilaterals like CUTS?Answer: Require that all the points of the quadrilateral, other than the vertices, lie on exactly one side. A crossing point lies on two sides, so CUTS is ruled out.
• Can you see how to rule out figures like OPENS?Answer: OPENS has five points and does not close up. Ruled out by saying a quadrilateral has exactly four vertices A, B, C, D and its sides are AB, BC, CD and DA, so the last side comes back to the first vertex.
• Note that quadrilateral ABCD can also be denoted as BCDA, CDAB, DABC, DCBA, ADCB, BADC or CBAD but not by any other sequence of vertices. (Draw ABCD and trace the vertices in various orders to see why.)Answer: A name must list the vertices in the order you meet them going round the boundary. You can start at any of the 4 vertices and go either way round: 4 × 2 = 8 names. Any other order (such as ACBD) joins A to C, a diagonal, and describes a different figure.
• Should DART in the figure be considered a quadrilateral? It seems different from our usual mental picture of a quadrilateral because it has a dent, as if someone has taken a bite out of triangle ART! We call DART a non-convex quadrilateral. A convex quadrilateral is one without a dent. How can we define this precisely?Answer: Yes, DART is a quadrilateral (a non-convex one): it meets every condition of the definition. A quadrilateral is convex when all of its internal angles are less than 180°; DART is non-convex because its internal angle at D is more than 180°.
• There are other ways of testing convexity. For example, visually verify that the diagonals of a convex quadrilateral intersect while those of a non-convex quadrilateral don’t intersect.Answer: In convex ABCD the diagonals AC and BD cross at a point inside. In non-convex DART, diagonal DR lies inside but diagonal AT lies outside the figure, and the two diagonal segments do not meet.
• Let ABCD be a quadrilateral.Answer: (i) Adjacent to AB: BC and DA; opposite to AB: CD. (ii) Adjacent to ∠A: ∠B and ∠D; opposite to ∠A: ∠C. (iii) Two sides are opposite if they have no common endpoint; two angles are opposite if their vertices are not the two ends of one side.
• You have used internal angles of quadrilaterals, but they too require an exact definition, just like how we gave one for a quadrilateral. Precisely define the internal angle of a quadrilateral at a given vertex. Your answer should work for a non-convex quadrilateral too. (Hint: use the opposite vertex as well.)Answer: In quadrilateral ABCD, the sides AB and AD make an angle ∠BAD (less than 180°). The internal angle at A is ∠BAD if A lies outside triangle BCD, and the reflex angle 360° − ∠BAD if A lies inside triangle BCD (C is the vertex opposite A).
• In a quadrilateral ABCD, suppose AB ‖ DC. Can ABCD be non-convex? What if we instead assume AB = CD? What if we instead assume ∠A = ∠C?Answer: If AB ‖ DC, no: ABCD must be convex. If AB = CD, yes, it can be non-convex (e.g. A(0, 0), B(6, 0), C(3, 1), D(3, 7)). If ∠A = ∠C, yes, it can be non-convex (an arrowhead with the dent at B or D).
• Consider three non-collinear points A, B, C and draw the lines AB, BC, CA. For every possible location of point D in the plane outside these lines, decide if ABCD is self-intersecting, non-convex, or convex. (Hint: the three lines divide the plane into 7 regions.)Answer: D inside ∆ABC: non-convex (dent at D). D across side CA from B: convex. D across side AB only, or across side BC only: self-intersecting. D in any of the three corner regions beyond a vertex: non-convex (dent at that vertex).
• Can a quadrilateral be both self-intersecting and non-planar?Answer: No. If two sides cross, say AB and CD meet at E, then lines AB and CD are two intersecting lines, and two intersecting lines lie in one plane. That plane contains A, B, C and D, so the quadrilateral is planar.
• To test if a given quadrilateral is a parallelogram, do we have to check that the opposite sides are parallel? Are there other ways to test this?Answer: No, we need not check parallel sides. A quadrilateral is a parallelogram if any one of these holds: both pairs of opposite sides equal; both pairs of opposite angles equal; the diagonals bisect each other; one pair of opposite sides both equal and parallel.
• Recall the following properties of a parallelogram that we proved last year.Answer: (a) ∆ACD ≅ ∆CAB (ASA), so AB = DC and AD = BC. (b) Adjacent angles are co-interior, so each pair adds to 180°; hence ∠A = ∠C and ∠B = ∠D. (c) ∆AED ≅ ∆CEB (ASA), so EA = EC and EB = ED.
• If the converse of any of the three properties is true, it can be used as an alternate way to show that a quadrilateral is a parallelogram. To explore this, let us write the converses. Can you experiment and guess what the answers are?Answer: All three converses are true: a quadrilateral is a parallelogram if (a) its opposite sides are equal, or (b) its opposite angles are equal, or (c) its diagonals bisect each other.
• If the opposite sides of a quadrilateral are of equal length, then it is a parallelogram.Answer: Join AC. ∆ACD ≅ ∆CAB by SSS (AB = CD, BC = DA, AC common). So ∠CAB = ∠ACD and ∠ACB = ∠CAD, which are alternate angles; hence AB ‖ DC and AD ‖ BC, and ABCD is a parallelogram.
• If the opposite angles of a quadrilateral are equal, then it is a parallelogram.Answer: ∠A + ∠B + ∠C + ∠D = 360° gives 2(x + y) = 360°, so ∠A + ∠B = ∠B + ∠C = 180°. Co-interior angles adding to 180° give BC ‖ AD (transversal AB) and AB ‖ DC (transversal BC).
• The angles of a quadrilateral add up to 360°. Therefore, if the opposite angles are equal, what can we say about adjacent angles? Is the converse of your answer true? Conclude that the result “if the opposite angles of a quadrilateral are equal, then it is a parallelogram” can also be stated as follows. “If each pair of adjacent angles in a quadrilateral ABCD …then ABCD is a parallelogram.” Fill in the blank.Answer: If opposite angles are equal, every pair of adjacent angles adds up to 180° (they are supplementary). The converse is true: if each pair of adjacent angles adds up to 180°, the opposite angles are equal. So: “If each pair of adjacent angles in a quadrilateral ABCD adds up to 180° (is supplementary), then ABCD is a parallelogram.”
• A quadrilateral whose diagonals bisect each other is a parallelogram.Answer: ∆AED ≅ ∆CEB by SAS (EA = EC, ∠AED = ∠CEB, ED = EB), so ∠DAE = ∠BCE and AD ‖ BC (alternate angles). Similarly ∆EAB ≅ ∆ECD gives AB ‖ DC. So ABCD is a parallelogram.
• A quadrilateral with one pair of equal and parallel opposite sides is a parallelogram.Answer: Let the diagonals meet at E. ∆EAB ≅ ∆ECD by ASA (∠EAB = ∠ECD, AB = CD, ∠EBA = ∠EDC, alternate angles). So EA = EC and EB = ED: the diagonals bisect each other, and ABCD is a parallelogram.
• Suppose in a quadrilateral ABCD we have AB ‖ DC and AB = DC. Let E be the intersection point of the diagonals AC and BD. (Why must the diagonals intersect?)Answer: Because AB ‖ DC makes ABCD convex: every vertex lies on one of two parallel lines with the whole figure in the strip between them, so no angle exceeds 180°. The diagonals of a convex quadrilateral always intersect.
• True or false?Answer: (i) True. (ii) False: every rhombus has perpendicular diagonals, but not every rhombus is a square. (iii) True.
• The diagonal AC of a parallelogram ABCD bisects ∠A. Show that it also bisects ∠C and that ABCD is a rhombus.Answer: Alternate angles give ∠ACD = ∠CAB and ∠ACB = ∠DAC. Since ∠CAB = ∠DAC, all four are equal, so ∠ACD = ∠ACB: AC bisects ∠C. In ∆ABC, ∠BAC = ∠BCA, so AB = BC; a parallelogram with two adjacent sides equal is a rhombus.
• The following questions examine converses of true properties. Answer them with Yes or No. If your answer is No, what extra condition can you add so that the answer becomes Yes?Answer: (i) Yes. (ii) Yes. (iii) No (an isosceles trapezium has equal diagonals); it becomes Yes if we add that the diagonals bisect each other (that is, ABCD is a parallelogram).
• Let ABCD be a parallelogram with AB ≠ BC. Show that the pairwise intersection points of the four angle bisectors form the vertices of a rectangle (see the figure). Why did we assume AB ≠ BC?Answer: Adjacent angles of a parallelogram add to 180°, so half of each adds to 90°. Hence the two bisectors from the ends of any side meet at 90°, so PQRS has four right angles: a rectangle. If AB = BC (a rhombus), the bisectors are the diagonals and all four points coincide at the centre, so there is no rectangle.
• On a piece of paper, draw and then cut out two copies of the same triangle. Keep one copy aside. On the other, mark the midpoint of each side. (You can do this by folding the paper to join two vertices at a time.) Draw the triangle made by the three midpoints and cut the paper along each of these lines. Now you have four smaller triangles. Compare them to each other and to the intact copy of the original triangle. What do you notice?Answer: The four smaller triangles are congruent to each other (they fit exactly on top of one another; the middle one has to be turned round). Each is a half-size copy of the original triangle: same angles, every side half as long, and one quarter of the area.
• Try to prove directly that the four smaller triangles are congruent. You will see that no congruence test applies, because many quantities are unknown.Answer: Comparing, say, ∆APQ and ∆PBR, we know only one pair of equal parts (AP = PB). We do not know AQ = PR, PQ = BR, or any equal angles, so SSS, SAS, ASA and RHS all fail. The missing facts (PQ = ½BC, QR = ½AB, RP = ½CA) come from the Midpoint Theorem, and then SSS works.
• The segment joining the midpoints of two sides of a triangle is parallel to the third side and has half the length.Answer: Draw line l through C parallel to BA, meeting line PQ at R. ∆APQ ≅ ∆CRQ (AAS), so PQ = QR and CR = AP = BP. Then BCRP has CR equal and parallel to BP, so it is a parallelogram: PQ ‖ BC and PQ = ½PR = ½BC.
• Read the statement of the Midpoint Theorem again. The following questions seem natural to ask. What can we say about (1) a segment PQ (with P on side AB and Q on side AC) that is parallel to the third side and has half the length, and (2) the line that passes through the midpoint of one side and is parallel to another side? For now, experiment and see if you can make a guess about question (2).Answer: (1) P and Q must be the midpoints of AB and AC. (2) The line through the midpoint of one side, parallel to a second side, bisects the third side (and the part inside the triangle is half the parallel side).
• The line drawn through the midpoint of one side of a triangle and parallel to another side bisects the third side (and the resulting segment has half the length of the parallel side).Answer: Draw the line through C parallel to BA, meeting line PQ at R. BCRP is a parallelogram, so CR = BP = PA. Then ∆APQ ≅ ∆CRQ (AAS), so AQ = QC (Q is the midpoint of AC) and PQ = QR, giving PQ = ½PR = ½BC.
• BCRP is a parallelogram, so CR = BP. BP = PA and ∠AQP = ∠CQR (vertically opposite angles). Combining with CR ‖ BA, ∆APQ ≅ ∆CRQ. (Why?)Answer: By AAS: AP = CR (both equal BP); ∠PAQ = ∠RCQ (alternate angles, CR ‖ BA with transversal AC); ∠AQP = ∠CQR (vertically opposite). (Equally, ∠APQ = ∠CRQ, alternate angles with transversal PR, gives ASA.)
• We will later generalise the converse of the Midpoint Theorem by dropping the condition that P is the midpoint of AB and assuming only that PQ ‖ BC. Can you guess the conclusion in this case? (Hint: the name of the more general theorem includes the word proportionality.)Answer: If PQ ‖ BC (P on AB, Q on AC), then AP : PB = AQ : QC, that is, APPB = AQQC. The line divides the two sides in the same ratio (the Basic Proportionality Theorem).
• In ∆ABC, suppose P, Q and R are the midpoints of side AB, AC and BC respectively. Instead of joining the midpoints with each other as we did earlier, draw segments AR, BQ and CP. These are called the medians of ∆ABC. What do you see? Repeat this for several triangles and see that the three medians always seem to be concurrent, meaning they pass through a common point! There is more! Assume for the moment that the medians are always concurrent. Let M be the common point where they meet. Compare the lengths of the two parts CM and MP of the median CP, and do the same for the other two medians. Do you see a pattern?Answer: The three medians always pass through one common point M (they are concurrent). On every median, the part from the vertex is twice the part from the midpoint: CM : MP = AM : MR = BM : MQ = 2 : 1.
• Let Y and X be the midpoints of CM and BM respectively. Now we can use the Midpoint Theorem in ∆ABC and in ∆MBC! We get that PQ and XY are parallel to BC and hence to each other. Moreover PQ = XY = BC2. So ∆MPQ ≅ ∆MYX by ASA (How?).Answer: PQ ‖ XY, so with transversal PY (the median CP) ∠MPQ = ∠MYX, and with transversal QX (the median BQ) ∠MQP = ∠MXY (alternate angles). The side between these angles is PQ = YX (both ½BC). So ∆MPQ ≅ ∆MYX by ASA.
• The three medians of ∆ABC pass through a common point, which divides each of the medians in the ratio 2 : 1, with the longer part connecting to the vertex. This point is called the centroid of ∆ABC.Answer: Let medians CP and BQ meet at M, and let X, Y be the midpoints of BM, CM. The Midpoint Theorem in ∆ABC and ∆MBC gives PQ ‖ XY and PQ = XY = ½BC, so ∆MPQ ≅ ∆MYX (ASA). Hence MQ = MX = XB and MP = MY = YC, i.e. BM : MQ = CM : MP = 2 : 1. The same argument for AR and BQ gives a point N with BN : NQ = 2 : 1; only one point divides BQ in this ratio, so N = M: all three medians pass through M.
• Give another proof of the Centroid Theorem using a clever construction that we will describe only partly and ask you to complete. Once again let M be the intersection point of the two medians CP and BQ. We want to show that the intersection point of line AM with side BC is the midpoint of BC. For this, extend AM up to a carefully chosen point S such that BSCM becomes a parallelogram. (Hint: To arrange MC ‖ BS, first note that PM and MC are along the same line. Now where should M be on segment AS? Does your placement of S also prove MB ‖ CS?)Answer: Choose S so that M is the midpoint of AS (MS = AM). In ∆ABS, P and M are midpoints, so PM ‖ BS, i.e. MC ‖ BS. In ∆ACS, Q and M are midpoints, so QM ‖ CS, i.e. MB ‖ CS. So BSCM is a parallelogram, its diagonals BC and MS bisect each other, and X (where AS meets BC) is the midpoint of BC. Hence the third median passes through M; also MX = ½MS = ½AM, so AM : MX = 2 : 1.
• The midpoints of the four sides of a quadrilateral are the vertices of a parallelogram.Answer: Midpoint Theorem in ∆ABC: PQ ‖ AC; in ∆ADC: SR ‖ AC; so PQ ‖ SR. In ∆BCD: QR ‖ BD; in ∆BAD: PS ‖ BD; so QR ‖ PS. Both pairs of opposite sides are parallel, so PQRS is a parallelogram (the Varignon parallelogram).
• (i) If P, Q, R are the midpoints of sides AB, AC, BC respectively of ∆ABC, show that ∆PQR is congruent to ∆QPA and to two other triangles which you should identify.
  (ii) Suppose someone erases ∆ABC, leaving only ∆PQR on the paper. Can you reconstruct ∆ABC from ∆PQR?Answer: (i) By the Midpoint Theorem PQ = ½BC, QR = ½AB, RP = ½AC, so by SSS ∆PQR ≅ ∆QPA ≅ ∆RBP ≅ ∆CRQ. (ii) Yes: through each vertex of ∆PQR draw the line parallel to the opposite side; these three lines form ∆ABC.
• In ∆ABC, let M and N be midpoints of AB and AC respectively. Let D be any point on BC. Show that MN bisects AD.Answer: MN ‖ BC (Midpoint Theorem). Let MN meet AD at E. In ∆ABD, the line through the midpoint M of AB parallel to BD meets AD at E, so E is the midpoint of AD (converse of the Midpoint Theorem): MN bisects AD.
• In a quadrilateral ABCD, suppose AB ‖ DC and AB ≠ CD. Suppose G and H are the midpoints of AC and BD respectively. Prove that GH ‖ AB. (Why did we assume AB ≠ CD?)Answer: Let K be the midpoint of AD. In ∆ADC, KG ‖ DC; in ∆DAB, KH ‖ AB ‖ DC. Only one line through K is parallel to DC, so K, H, G are collinear and GH ‖ AB. If AB = CD, ABCD would be a parallelogram, whose diagonals bisect each other, so G = H and there is no segment GH.
• Suppose the midpoints of sides AB, BC, CD and DA of a quadrilateral ABCD are P, Q, R and S respectively.Answer: (i) PQRS is a parallelogram (midpoints of a quadrilateral), and PR, QS are its diagonals, so they bisect each other. (ii) If AC = BD then PQ = ½AC = ½BD = QR, so PQRS is a rhombus and its diagonals PR ⟂ QS. The converse is true: PR ⟂ QS makes the parallelogram PQRS a rhombus, so ½AC = ½BD, i.e. AC = BD.
• Suppose PQRS is the Varignon parallelogram of ABCD.Answer: (i) Place A′ so that ∆SA′P ≅ ∆SAP (copy SA and PA, same side as A). Extend A′P to B′ with PB′ = A′P, B′Q to C′ with QC′ = B′Q, C′R to D′ with RD′ = C′R; join D′A′. (ii) Step by step ∆PB′Q ≅ ∆PBQ, ∆QC′R ≅ ∆QCR, ∆RD′S ≅ ∆RDS (SAS); the angles at S add to 180°, so A′, S, D′ are collinear and the four corner triangles and PQRS match those of ABCD: A′B′C′D′ ≅ ABCD. (iii) PQ = ½AC, QR = ½BD, PQ ‖ AC, QR ‖ BD: PQRS a square ⇔ AC = BD and AC ⟂ BD.
• Suppose we have a tiling of the plane. Consider any vertex. As we go around this vertex and consider the angles made by consecutive lines, the total of these angles must be 360°. This suggests an idea. What if we take 4 copies of SOME and fit them together around a common point so that each angle is used once as we go around?Answer: It works: angles 1 + 2 + 3 + 4 = 360° (the angle sum of a quadrilateral), so four copies with one of each angle at the point fill the full turn exactly, with no gap and no overlap around that point. There are three essentially different orders in which the four angles can go round.
• There are multiple ways of doing this, as shown in the figure. Can you use any of these ways to continue fitting further copies of SOME to tile the plane? Try it with the 15 copies you made!Answer: Yes. Use the arrangement in which the angles go round the point in the order 1, 2, 3, 4. In it each copy is the neighbouring copy turned through 180° about the midpoint of their common edge, so neighbours share whole edges and the pattern can be extended in every direction.
• How can we understand the figure? There seems to be a repeating pattern. (1) Can you precisely describe a procedure to draw the pattern so that someone can draw the tiling on their own, based only on your description? (2) Can you justify why your procedure works?Answer: (1) Draw SOME. Turn a copy through 180° about the midpoint of OM to get its partner. Then copy this pair again and again, sliding it by the diagonal SM (forwards and backwards) and by the diagonal OE, in every combination. (2) The half-turn makes the pair share edge OM exactly; the slides by the diagonals make the remaining edges meet edge to edge; at every corner the four angles 1, 2, 3, 4 meet once, adding to 360°, so there are no gaps or overlaps.
• Note two interesting things about the second step: (1) each new copy can be obtained by rotating any one of its neighbours, and both ways give the same result. (2) The new copies fit perfectly. Can you explain these facts by reasoning? This is needed to prove that the method works!Answer: (1) Two half-turns about different points make a slide by twice the distance between the points. Reaching a corner copy through either neighbour gives the same slide (by the diagonal of SOME from the start tile’s vertex to the corner), so both ways give the same copy. (2) That copy brings the missing angle to the corner, so the four angles there add to 360°, and its two edges at the corner coincide with the neighbours’ edges, so it fits perfectly.
• Do you see which of the three possibilities shown in the figure occurs in the tiling?Answer: The third possibility: going round each corner the angles appear in the order 1, 2, 3, 4 (read one way or the other), so angle 1 is opposite angle 3 and angle 2 is opposite angle 4.
• A grid of Varignon parallelograms of SOME is given in the figure. To begin with, focus only on the 9 green coloured copies of SOME in the figure. We will first place only these 9 copies in the figure. The corresponding parallelograms are shaded in the figure. Place 9 copies of SOME so as to match the way the green coloured copies of SOME are placed in the figure. If done carefully, you will observe that each 4-gon you placed meets other placed copies exactly at vertices. Now see how 4 copies of SOME are made automatically in the gaps! Carefully place four more copies of SOME in these gaps. Continuing this process with more copies of SOME will give us the desired tiling.Answer: Place each copy so that the midpoints of its sides are the corners of a shaded cell. The 9 copies are all slid copies of SOME and touch only at corners. Each gap is bounded by four sides of four different copies (one side equal to each side of SOME) and is exactly a copy of SOME turned through 180°; filling the gaps and repeating gives the tiling.
• The following shape— the first of its kind!— that leads to aperiodic tilings was discovered only recently in 2023 by a team of four mathematicians (Smith, Myers, Kaplan and Goodman-Strauss). Cut out 15 identical copies of the following shape (known as the ‘hat’) and start to tile with it. Do you see any pattern?Answer: The hats fit together without gaps (some copies must be turned over, as mirror images), but no regular repeating pattern appears: there is no block that you can simply slide again and again to cover the plane. The hat tiles the plane only aperiodically.
• Justify why the plane cannot be tiled with a regular pentagon. (Hint: Read the first 3 sentences of ‘Think and Reflect’ in the section on tiling.) (There are many ways to tile the plane using a suitable irregular pentagon. The most recent method was found in 2015.)Answer: Each angle of a regular pentagon is 108°. Round a corner point of a tiling the angles must total 360°, but 360 ÷ 108 = 3⅓ is not a whole number (3 corners give 324°, 4 give 432°). A corner touching the middle of another tile’s edge would need 180° from corners, and 180 ÷ 108 is not whole either. So regular pentagons cannot tile the plane.
• Draw a non-convex 4-gon DART. Show how we can tile the plane with copies of DART. Both methods that we discussed earlier will work. Which do you prefer?Answer: Number the angles of DART 1, 2, 3, 4 (the reflex angle at D is one of them; they still add to 360°). Method 1: turn copies through 180° about the midpoints of the sides, again and again. Method 2: draw the grid of midpoint parallelograms of DART, place copies on every other cell, and fill the gaps with half-turned copies. Both tile the plane. Method 1 is easier to carry out; Method 2 shows the structure more clearly (either preference is fine with a reason).
• Using a fact about parallelograms, show how to tile the plane using any given triangle. (Hint: Can you use the parallelogram tiling in the introduction?)Answer: Turn a copy of ∆ABC through 180° about the midpoint M of BC; call the image of A, A′. The diagonals AA′ and BC of ABA′C bisect each other at M, so ABA′C is a parallelogram. Copies of a parallelogram tile the plane (the grid of parallel lines), so splitting each parallelogram back into its two triangles tiles the plane with copies of ∆ABC.
• Mark the midpoint of the line drawn on the paper (see the figure), given that the horizontal lines are equally spaced. Justify your answer.Answer: The segment goes from one ruled line (X) to the line four spaces above it (Y). Its midpoint is the point K where it crosses the middle line, two spaces from each end. Reason: take Z on Y’s line straight above X; the middle line passes through the midpoint W of XZ and is parallel to ZY, so by the converse of the Midpoint Theorem it bisects XY.
• You know that the sum of angles of a quadrilateral is 360°, even for a non-convex quadrilateral. (Recall the proof.) Now consider a self-intersecting quadrilateral ABCD, where AB and CD intersect at point E. Show that ∠A + ∠B + ∠C + ∠D < 360°. Can you construct ABCD such that ∠A + ∠B + ∠C + ∠D = 2°?Answer: With ∠BEC = ∠AED = θ (vertically opposite): in ∆EBC, ∠B + ∠C = 180° − θ; in ∆EAD, ∠A + ∠D = 180° − θ. So ∠A + ∠B + ∠C + ∠D = 360° − 2θ < 360°. Yes, sum = 2° when θ = 179°: e.g. E at the centre, EA = EB = EC = ED, with rays EB and EC making 179° (and A, D opposite B, C). Then each angle of ABCD is 0.5°.
• In a parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see the figure). Show that APCQ is a parallelogram.Answer: Let the diagonals AC and BD meet at O. Then OA = OC and OB = OD. So OP = OD − DP = OB − BQ = OQ. The diagonals AC and PQ of APCQ bisect each other at O, so APCQ is a parallelogram.
• A right-triangle shaped cutout of a paper is folded such that point A touches point B (see the figure). Show that the crease line can be used to find the midpoint of not only AB but also that of AC.Answer: Folding A onto B makes the crease the perpendicular bisector of AB: it passes through the midpoint M of AB and is ⟂ AB. Since BC ⟂ AB too, the crease is parallel to BC. By the converse of the Midpoint Theorem, the line through M parallel to BC bisects AC, so the crease meets AC at its midpoint N.
• You saw how to use the Midpoint Theorem to divide a given triangle into 4 congruent triangles. Can we divide a triangle into 3 congruent triangles? This exercise shows us how to do that and more, provided we are allowed to cut and reassemble.Answer: (i) Turn ∆MPB through 180° about P (B goes to C) and join it to ∆MPC: since ∠MPB + ∠MPC = 180°, the result is a triangle MCM′ with sides ⅔AP, ⅔BQ, ⅔CR. Pairing the pieces at Q and at R the same way gives two more triangles with the same sides, so the 3 are congruent (SSS). (ii) Each assembled triangle has medians ½BC, ½CA, ½AB, so repeating (i) gives triangles with sides ⅔ × ½ of these: AB3, BC3, AC3. (iii) Cut along median AP, then cut ∆APB along its median PX (X the midpoint of AB); turn ∆PXB about X onto A: it forms a triangle with sides AP, PC, CA, congruent to ∆APC.
• A more general midpoint theorem and its converse. In a quadrilateral ABCD, suppose AB ‖ DC. Recall that such ABCD is called a trapezium. Let E be the midpoint of AD. A line drawn through E intersects side BC at F.Answer: (i) Draw BD meeting EF at G. In ∆ABD, the line through midpoint E parallel to AB bisects BD (G is its midpoint) and EG = ½AB. In ∆BDC, the line through G parallel to DC bisects BC, so F is the midpoint of BC, and GF = ½DC. So EF = AB + CD2. (ii) With M the midpoint of BD, EM ‖ AB (in ∆DAB) and FM ‖ DC ‖ AB (in ∆BCD); only one line through M is parallel to AB, so E, M, F are collinear and EF ‖ AB.
• The diagonals AC and BD of a parallelogram ABCD intersect at O. A line through O meets AB and CD at points P and Q respectively. Show that O is the midpoint of PQ. (Multiple proofs are possible. Which is the simplest?)Answer: In ∆OAP and ∆OCQ: OA = OC (diagonals bisect each other), ∠OAP = ∠OCQ (alternate angles, AB ‖ DC), ∠AOP = ∠COQ (vertically opposite). So ∆OAP ≅ ∆OCQ (ASA) and OP = OQ: O is the midpoint of PQ. This congruence proof is the simplest.
• ABCD is a trapezium with parallel sides AD = 3 cm and BC = 5 cm. E and F are the midpoints of the non-parallel sides. Find the ratio of the areas of the 4-gons AEFD and EBCF.Answer: EF ‖ AD ‖ BC and EF = ½(3 + 5) = 4 cm. EF is halfway between AD and BC, so both parts have height h/2. ar(AEFD) : ar(EBCF) = ½(3 + 4)·h2 : ½(4 + 5)·h2 = 7 : 9.
• Consider 4 points A, B, C, D in the plane with no three collinear. Answer the following questions. Some answers may require you to consider different cases, depending on how the points are positioned in the plane.Answer: (i) 3 quadrilaterals: ABCD, ABDC, ACBD. (ii) If none of the points lies inside the triangle formed by the other three: 2 self-intersecting, 1 convex. If one point lies inside the triangle of the other three: 0 self-intersecting, 0 convex (all 3 are non-convex).
• Suppose P is a point on side AB of ∆ABC and the line through P parallel to BC meets AC in point Q. For parts (i) to (iii), assume AP = 1, AQ = √5, and see the figure.Answer: (i) QC = 2√5. (ii) QC = √53. (iii) QC = 2√53. (iv) If AP : PB = m : n, cut AB into m + n equal parts and draw parallels to BC through the cut points; they cut AC into m + n equal parts, so AQ : QC = m : n = AP : PB.
• (i) Suppose ABCD is a parallelogram and M, N are midpoints of AB and CD respectively. Show that segments DM and BN trisect segment AC.
  (ii) Use part (i) to find a procedure to trisect any given segment PQ. Find two other ways to trisect PQ, one using the fact that a line parallel to one side of a triangle divides the other two sides in the same ratio and a third using the Centroid Theorem.Answer: (i) MBND is a parallelogram, so DM ‖ BN. With X = DM ∩ AC and Y = BN ∩ AC: in ∆ABY, MX ‖ BY through the midpoint M, so AX = XY; in ∆CDX, NY ‖ DX through the midpoint N, so CY = YX. Hence AX = XY = YC. (ii) Three ways: (a) build a parallelogram with diagonal PQ and use (i); (b) mark 3 equal steps on any ray from P, join the last to Q and draw parallels; (c) make PQ a median of a triangle: the centroid is ⅔ of the way from P.
• Is the Midpoint Theorem for Quadrilaterals true when the quadrilateral is non-convex? How about when it is self-intersecting? Experiment and check. The pictures may look strange, but the theorem appears to be true.Answer: Yes, in both cases the midpoints form a parallelogram. (i) The proof uses only the Midpoint Theorem in the four triangles ABC, ADC, BCD, BAD, which exist however ABCD is shaped: PQ ‖ AC ‖ SR and QR ‖ BD ‖ PS. (ii) It becomes a segment exactly when AC ‖ BD (possible only for a self-intersecting ABCD), because then all four sides of PQRS lie along one direction. (iii) Yes: the four triangles still exist, PQ and SR are both parallel to AC, so P, Q, R, S lie in one plane and PQRS is a parallelogram.
• Let P, Q, R, S be four points on sides AB, BC, CD and DA respectively of a quadrilateral ABCD. Suppose PQRS is a parallelogram. Must P, Q, R and S be midpoints of the respective sides? This can be considered a possible converse question to the Midpoint Theorem for Quadrilaterals.Answer: No. In a square ABCD of side 4, take AP = BQ = CR = DS = 1. The four corner triangles are congruent (SAS), so PQ = QR = RS = SP: PQRS is a rhombus (in fact a square), hence a parallelogram, but none of P, Q, R, S is a midpoint.
• (i) Complete the following proof of the Midpoint Theorem. Extend segment PQ beyond Q until point S. By how much should we extend PQ? It would be good to be able to prove that ∆APQ ≅ ∆CSQ. (Why?) Use this as a guide to specify the location of S and then complete this proof.
  (ii) Give a similar proof of the converse of the Midpoint Theorem. Start by extending PQ up to a suitable point S.Answer: (i) Extend PQ by its own length: QS = PQ. Then ∆APQ ≅ ∆CSQ (SAS), so CS = AP = PB and CS ‖ AB (∠PAQ = ∠SCQ, alternate). So PBCS is a parallelogram: PQ ‖ BC and PQ = ½PS = ½BC. The congruence is wanted because it moves AP across to CS, where it is equal and parallel to PB. (ii) For the converse (P the midpoint, PQ ‖ BC), extend PQ to S with PS = BC: PBCS is a parallelogram, so CS = PB = AP and CS ‖ AB; then ∆APQ ≅ ∆CSQ (ASA) gives AQ = QC (and PQ = QS = ½BC).
• Review all the properties of a rhombus/rectangle/square that you proved earlier. Formulate a converse of each. Decide if the converse is true. There are many possibilities here!Answer: Sample results. True converses: a quadrilateral whose diagonals bisect each other at right angles is a rhombus; whose diagonals bisect each other and are equal is a rectangle; whose diagonals are equal and bisect each other at right angles is a square; whose diagonals bisect all four angles is a rhombus; with all angles 90° is a rectangle. False converses: “equal diagonals ⇒ rectangle” (isosceles trapezium); “perpendicular diagonals ⇒ rhombus” (kite); “equal and perpendicular diagonals ⇒ square” (a kite with equal diagonals).
• Show that the sum of angles of a non-planar quadrilateral is always less than 360°. Can you find a non-planar quadrilateral ABCD for which ∠A + ∠B + ∠C + ∠D = 2°? (Hint: Think of a diagonal, say AC, as a hinge around which triangles ABC and ADC can rotate.) What happens to each angle of ABCD as you do this rotation?Answer: Hinge triangles ABC and ADC on AC. ∠B and ∠D never change. Flat (planar, D away from B), ∠A = ∠BAC + ∠CAD and ∠C = ∠BCA + ∠ACD, so the sum is 180° + 180° = 360°. Folding moves D closer to B, which makes ∠A = ∠BAD and ∠C = ∠BCD smaller; so any non-planar ABCD has sum < 360°. Yes, 2° is possible: take AB = CB with ∠ABC = 0.5°, D the mirror copy of B, and fold until the two triangles make an angle of 0.5°; then the sum is 2°.
• Let us see a third method to tile the plane using a 4-gon. Focus on only two coloured copies of SOME sharing a vertex. What do you see? It appears that each 4-gon is just a shifted copy of the other. Let us see exactly how. Draw two copies of SOME as shown in the figure so that the diagonals EO and EʹOʹ are collinear with O = Eʹ. Now place a cutout of one copy on top of SOME with diagonal EO drawn on it. Slide this cutout so that segment EO moves along EOʹ until EO matches EʹOʹ. Verify that your cutout exactly matches SʹOʹMʹEʹ.Answer: (i) S′O′M′E′ is a copy of SOME with diagonal E′O′ on the line EO, same length and same direction, and S′, M′ on the same sides. So ∠SEO = ∠S′E′O′ (corresponding angles) gives ES ‖ E′S′, and ES = E′S′: ESS′E′ is a parallelogram. In the same way EMM′E′, OSS′O′, OMM′O′ are parallelograms. So SS′, MM′, EE′ and OO′ are all equal and parallel to EO: every vertex moves by the same slide, and the cutout matches S′O′M′E′ exactly. (ii) Sliding along both diagonals both ways gives a 3 × 3 grid of copies that meet only at corners; each blank space has sides equal to SO, OM, ME, ES and angles 1, 2, 3, 4, so it is a copy of SOME turned through 180°.
• Is there a 4-gon with given side lengths?Answer: (i) Draw AB = c, then arcs of radius b from A and a from B: they meet (giving a triangle) exactly when a + b > c (with a ≤ b ≤ c). (ii) 100: no; 10: yes; 1: no. Possible fourth sides: 4 < x < 18. (iii) A 4-gon exists exactly when the longest side is less than the sum of the other three.
• Counting diagonals of a polygon.Answer: A diagonal joins two vertices that are not adjacent. Diagonals for n = 3, 4, 5, 6, 7, 8: 0, 2, 5, 9, 14, 20. Formula: n(n − 3)2. Going from n to n + 1 sides adds n − 1 diagonals.
• Sum of angles of a polygon. What is the sum of angles of a (planar non-self-intersecting) n-gon? We know that the answer is 180° for n = 3 and 360° for n = 4. Find the next few values. Then find a formula in terms of n and prove it.Answer: n = 5: 540°; 6: 720°; 7: 900°; 8: 1080°. Formula: (n − 2) × 180°. Proof: diagonals lying inside the polygon cut it into n − 2 triangles, whose angles together make up exactly the angles of the polygon.
• Multiple converses to a theorem. Let us see how multiple statements can be considered converses to the Midpoint Theorem and how the converse of the Midpoint Theorem (the line through the midpoint of one side parallel to another side bisects the third side) is one of them. To formulate a converse we should express the original statement in “If … then …” form. For a complex statement, there may be multiple ways to do that. The Midpoint Theorem starts with ∆ABC and points P and Q on sides AB and AC respectively. The theorem has two assumptions and two conclusions, which we have named for further discussion.
  Assumptions: (P MID) P is the midpoint of AB, and (Q MID) Q is the midpoint of AC.
  Conclusions: (PRLL) PQ ‖ BC, and (HALF) PQ = BC2.
  Let us use these four named conditions to discuss various possible statements.Answer: (i) “In ∆ABC with P on AB and Q on AC, if PQ ‖ BC and PQ = ½BC, then P and Q are the midpoints of AB and AC.” True. (ii) “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 the converse of the Midpoint Theorem, i.e. (P MID) and (PRLL) ⇒ (Q MID) (and (HALF)). (iii) The only remaining pair is (P MID) and (HALF); no, neither (Q MID) nor (PRLL) need follow: e.g. A(0, 0), B(3, 3), C(4, 0), P(1.5, 1.5) and Q(1, 0) has PQ = ½BC but Q is not the midpoint and PQ ∦ BC.
• Validity of tiling methods. Show using reasoning that each of the three tiling methods we saw produces a tiling of the plane using the given 4-gon. You have to prove that when we place new copies using any of the three procedures, the entire plane is covered with no gaps and no overlaps. First, think carefully about what it means to prove this. Then find a proof for each procedure.Answer: What to prove: every point of the plane lies in some copy (no gaps), and no two copies share inside points (no overlaps). Method 1: SOME and its half-turn about the midpoint of OM form a hexagon whose opposite sides are equal and parallel; such a hexagon tiles by slides (along the diagonals SM and OE) row by row, exactly like a parallelogram, and its copies are the copies placed by Method 1. Method 2: the Varignon cells form a parallelogram grid that covers the plane; each placed copy covers its own cell plus four corner triangles, and around every grid vertex-cell the four corner triangles of the four surrounding copies fill that cell exactly. Method 3: it places the same slid copies as Method 1 (slides along the diagonals), with the half-turned copies in the gaps, so it gives the same tiling.
• What fraction of the square is shaded?Answer: 15. Let the inner square have side s. Midpoint arguments give DP = PQ = s and CQ = QR = s, so in the right triangle DQC, DQ = 2s and QC = s. By Pythagoras DC² = 4s² + s² = 5s², so the shaded square is s²5s² = 15 of the whole.

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