NCERT Solutions Class 9 Mathematics Ganita Manjari 2 Chapter 12 Quadrilaterals End of Chapter Exercises

End-of-Chapter Exercises

Question 1. 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:
Let ABC be the triangle and M the midpoint of BC.
Step 1: two copies make a parallelogram. Turn a copy of △ABC through 180° about M. B and C swap places and A goes to A′, with M the midpoint of AA′. In ABA′C the diagonals AA′ and BC bisect each other at M, so by Theorem 4 ABA′C is a parallelogram made of △ABC and △A′CB.

Step 2: parallelograms tile the plane. Place copies side by side, sliding along one side, to make a strip; then stack copies of the strip, sliding along the other side. Opposite sides of a parallelogram are equal and parallel, so the strips fit exactly.
Step 3. Cut every parallelogram along the same diagonal BC. Each splits into two copies of △ABC, so the plane is tiled by copies of the triangle.

Question 2. Mark the midpoint of the line drawn on the paper (see Fig. 12.33), given that the horizontal lines are equally spaced. Justify your answer.

Answer:
The segment runs from X on one ruled line to Y on the ruled line four spaces higher. Call the line through X l₁ and the line through Y l₅; the lines in between are l₂, l₃, l₄.
Answer: the midpoint M of XY is where XY crosses the middle ruled line l₃ (two spaces above X and two below Y).

Justification. Through X draw XZ perpendicular to the ruled lines, meeting l₅ at Z. The lines are equally spaced, so l₃ is two spaces from both l₁ and l₅ and meets XZ at its midpoint N: XN = NZ.
In △XZY, N is the midpoint of XZ and l₃ (through N) is parallel to ZY, since both are ruled lines. By Theorem 7, l₃ bisects the third side XY. So XM = MY.
(In general, equally spaced parallel lines cut equal lengths on any line crossing them.)

Question 3. 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:
Recall: for an ordinary quadrilateral a diagonal splits it into two triangles whose angles make up the four angles, giving 2 × 180° = 360°.
Self-intersecting case. E lies on both AB and CD, and the figure is made of △EBC and △EAD. The angles are ∠A = ∠DAE, ∠B = ∠EBC, ∠C = ∠BCE and ∠D = ∠EDA.
From the angle sums of the two triangles: ∠B + ∠C = 180° − ∠BEC and ∠A + ∠D = 180° − ∠AED.
∠AED = ∠BEC (vertically opposite). Adding: ∠A + ∠B + ∠C + ∠D = 360° − 2∠BEC.
Since ∠BEC > 0°, ∠A + ∠B + ∠C + ∠D < 360°.
Sum equal to 2°. We need 360° − 2∠BEC = 2°, i.e. ∠BEC = 179°. Yes, it can be done:
Draw a line and mark E on it, with A and B on opposite sides of E. Draw a second line through E making 179° with EB (only 1° with EA). Mark C on it on the same side as A and D on the other side, near B. Join BC and DA.
Then AB and CD cross at E with ∠BEC = ∠AED = 179°, and ∠A + ∠B + ∠C + ∠D = 360° − 2 × 179° = 2°. (The figure is very thin, so it is drawn not to scale.)

Teacher's Note:
1. Each triangle loses the angle at E, so twice ∠BEC is lost from 360°.
2. Making ∠BEC close to 180° makes the sum as small as we like.

Question 4. In a parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 12.34). Show that APCQ is a parallelogram.

Answer:
Join AC and let the diagonals AC and BD meet at O. The diagonals of a parallelogram bisect each other, so OA = OC and OB = OD.
Given DP = BQ. Then OP = OD − DP = OB − BQ = OQ.
In quadrilateral APCQ the diagonals are AC and PQ, and they meet at O with OA = OC and OP = OQ.
So the diagonals of APCQ bisect each other, and by Theorem 4 APCQ is a parallelogram.

Question 5. A right-triangle shaped cutout of a paper is folded such that point A touches point B (Fig. 12.35). Show that the crease line can be used to find the midpoint of not only AB but also that of AC.

Answer:
In Fig. 12.35, ∠B = 90°. Let the crease meet AB at M and AC at N.
Crease and AB. When A is folded onto B, every point on the crease is equally far from A and B, so the crease is the perpendicular bisector of AB. So M is the midpoint of AB and MN ⊥ AB.
Crease and AC. BC ⊥ AB as well (right angle at B). Two lines perpendicular to the same line are parallel, so MN ∥ BC.
In △ABC, MN passes through the midpoint M of AB and is parallel to BC. By Theorem 7 it bisects AC, so N is the midpoint of AC.
So the crease marks the midpoint M of AB and the midpoint N of AC (and MN = \( \frac{1}{2} \)BC).

Question 6. 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.
(i) Draw medians AP, BQ and CR of △ABC, meeting in a common point M. Cut along each median to get 6 triangles. Show with justification how to assemble the 6 pieces into 3 congruent triangles! (Hint: Align △MPB and △MPC along equal sides PB and PC, matching P with itself and B with C.) Prove that you get a triangle. What are its side lengths?
*(ii) Repeat the procedure with each of the 3 assembled triangles you got in (i). Show that each of the resulting 9 triangles has sides \( \frac{AB}{3} \), \( \frac{BC}{3} \) and \( \frac{AC}{3} \).
The procedure and the result above were unnoticed till 2014, when they were discovered by Lee Sallows, an amateur mathematician!
(iii) Naturally, we can next ask about the possibility of cutting a triangle and reassembling it into 2 congruent triangles. You were asked this question in the chapter Measuring Space: Perimeter and Area. Can you solve this now? (Hint: Use median AP in △ABC and then a median of △APB.)
For any two given figures with straight sides, we can ask whether one figure can be cut (using only straight cuts) into pieces and reassembled to make the other. This notion is called scissors congruence. The figures need not be in the plane. Research on such questions at a higher level is ongoing.
Answer:

(i) P, Q, R are the midpoints of BC, CA, AB and, by the Centroid Theorem, AM = 2MP, BM = 2MQ, CM = 2MR.
Joining △MPB and △MPC. Turn △MPB through 180° about P. As P is the midpoint of BC, B goes to C, P stays, and M goes to M′ with P the midpoint of MM′. The turned piece △M′PC lies along PC next to △MPC.
It is a triangle. At P the two pieces have angles ∠MPC and ∠M′PC = ∠MPB. Since B, P, C are collinear, ∠MPC + ∠MPB = 180°, so M, P, M′ are on a straight line and the two pieces form △MCM′.

Side lengths: MM′ = 2MP = AM, MC = CM and CM′ = BM. So the triangle has sides AM, BM, CM, that is \( \frac{2}{3} \)AP, \( \frac{2}{3} \)BQ, \( \frac{2}{3} \)CR (two-thirds of the medians).
The other two. Turning △MQA about Q (A goes to C) and joining it to △MQC gives a triangle with sides 2MQ = BM, MC, MA. Turning △MRB about R (B goes to A) and joining it to △MRA gives sides 2MR = CM, MA, MB.
All three triangles have sides AM, BM, CM, so they are congruent (SSS). The 6 pieces form 3 congruent triangles.
(ii) Take T = △MCM′ from (i) and find its medians. BM′CM is a parallelogram (its diagonals BC and MM′ bisect each other at P), so CM′ ∥ BM and CM′ = BM.
1. Median from C: P is the midpoint of MM′, so CP is a median of T and CP = \( \frac{1}{2} \)BC.
2. Median from M: let K be the midpoint of CM′. Then CK = \( \frac{1}{2} \)CM′ = \( \frac{1}{2} \)BM = MQ, and CK ∥ BM, which lies along BQ. So in QMKC the sides QM and KC are equal and parallel (both pointing from M towards B), and QMKC is a parallelogram (Theorem 5). Hence MK = QC = \( \frac{1}{2} \)AC.
3. Median from M′: let L be the midpoint of MC. R, M, L, C lie on median CR and RL = RM + ML = \( \frac{1}{2} \)MC + \( \frac{1}{2} \)MC = MC (using CM = 2MR). From the parallelogram BM′CM, BM′ = MC and BM′ ∥ MC, in the same direction as RL. So RLM′B is a parallelogram and M′L = RB = \( \frac{1}{2} \)AB.
So the medians of T are \( \frac{1}{2} \)BC, \( \frac{1}{2} \)AC and \( \frac{1}{2} \)AB. By part (i), the 6 pieces of T form 3 congruent triangles with sides two-thirds of these: \( \frac{2}{3} \) × \( \frac{1}{2} \)AB = \( \frac{AB}{3} \), and similarly \( \frac{BC}{3} \), \( \frac{AC}{3} \).
The three triangles from (i) are congruent, so each behaves the same way. Altogether we get 3 × 3 = 9 triangles, each with sides \( \frac{AB}{3} \), \( \frac{BC}{3} \), \( \frac{AC}{3} \): a one-third copy of △ABC.
(iii) Cut along the median AP (P the midpoint of BC), then cut △APB along its median PR (R the midpoint of AB). The pieces are I = △APC, II = △RBP and III = △RAP.

Turn piece III through 180° about R: A goes to B and P goes to P′, with R the midpoint of PP′. Since ∠BRP + ∠ARP = 180°, P, R, P′ are collinear, so pieces II and III together form △BPP′.
Its sides: BP = \( \frac{1}{2} \)BC, BP′ = AP (the image of AP), PP′ = 2PR = AC (Midpoint Theorem in △ABC gives PR = \( \frac{1}{2} \)AC).
Piece I, △APC, has sides PC = \( \frac{1}{2} \)BC, AP and AC. So △BPP′ ≅ △CPA (SSS), and △ABC has been cut into three pieces that reassemble into two congruent triangles.

Teacher's Note:
1. Every reassembly here is a half-turn about a midpoint, which carries one end of a side onto the other.
2. Check the straight-line condition (angles adding to 180°) before calling the result a triangle.
3. Part (ii) is starred; the key result is that the medians of T are half the sides of △ABC.

Question 7. 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.
(i) If EF ∥ AB, then show that F is the midpoint of BC. Conclude that EF = \( \frac{AB + CD}{2} \).
(ii) If F is the midpoint of BC, then show that EF ∥ AB.
There are at least two ways to solve (ii). Do it directly by using the midpoint M of BD and showing that EM and FM are the same lines. Or use (i) along with the same idea used in the second proof of the converse of the Midpoint Theorem. Which way do you think is simpler?
Answer:
(i) Join AC, meeting EF at G. As EF ∥ AB and AB ∥ DC, also EF ∥ DC.
1. In △ADC: E is the midpoint of AD and EG ∥ DC. By Theorem 7, G is the midpoint of AC and EG = \( \frac{1}{2} \)DC.
2. In △CAB: G is the midpoint of CA and GF ∥ AB. By Theorem 7, F is the midpoint of CB and GF = \( \frac{1}{2} \)AB.
So F is the midpoint of BC, and EF = EG + GF = \( \frac{1}{2} \)DC + \( \frac{1}{2} \)AB = \( \frac{AB + CD}{2} \).
(ii) First way (direct). Let M be the midpoint of BD.

1. In △DAB: E and M are midpoints of DA and DB, so EM ∥ AB.
2. In △BDC: M and F are midpoints of BD and BC, so MF ∥ DC, i.e. MF ∥ AB.
3. EM and MF are both lines through M parallel to AB; there is only one such line, so E, M, F are collinear and EF ∥ AB.
Second way. Draw the line through E parallel to AB, meeting BC at F′. By (i), F′ is the midpoint of BC. F is also the midpoint of BC, and a segment has only one midpoint, so F′ = F and EF ∥ AB.
Which is simpler? The second way is shorter, as it reuses (i) with the uniqueness of the midpoint (the idea of the second proof of Theorem 7). The first way is also short and does not need (i).

Teacher's Note:
1. G (on AC) is the bridge between the two triangles in part (i).
2. EF is the average of the parallel sides; a quick numerical check helps students remember it.

Question 8. 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:
Proof 1 (congruent triangles). In △OAP and △OCQ:
1. OA = OC (diagonals of a parallelogram bisect each other);
2. ∠AOP = ∠COQ (vertically opposite angles);
3. ∠OAP = ∠OCQ (alternate angles, AB ∥ DC with transversal AC).
So △OAP ≅ △OCQ (ASA), giving OP = OQ: O is the midpoint of PQ.
Proof 2 (parallelogram test). Take Q′ on line PO beyond O with OQ′ = OP. The diagonals AC and PQ′ of APCQ′ bisect each other at O, so APCQ′ is a parallelogram (Theorem 4) and CQ′ ∥ AP, i.e. CQ′ ∥ AB ∥ CD. So Q′ is on line CD. Q′ is also on line PO, hence Q′ = Q and OP = OQ.
Simplest: Proof 1, which needs only one congruence.

Question 9. 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:
E and F are the midpoints of AB and DC. By Question 7(i), EF ∥ AD ∥ BC and EF = \( \frac{AD + BC}{2} \) = \( \frac{3 + 5}{2} \) = 4 cm.
Heights. Let h be the distance between AD and BC. Draw a perpendicular from a point of AD to BC. EF is parallel to BC and passes through the midpoint of AB, so (Theorem 7) it also bisects this perpendicular. So AEFD and EBCF each have height \( \frac{h}{2} \).
Areas (area of trapezium = \( \frac{1}{2} \) × sum of parallel sides × height):
ar(AEFD) = \( \frac{1}{2} \) × (3 + 4) × \( \frac{h}{2} \) = \( \frac{7h}{4} \); ar(EBCF) = \( \frac{1}{2} \) × (4 + 5) × \( \frac{h}{2} \) = \( \frac{9h}{4} \).
Ratio = \( \frac{7h}{4} \) : \( \frac{9h}{4} \) = 7 : 9.

Teacher's Note:
1. Name the vertices carefully: here the parallel sides are AD and BC.
2. The height h cancels, so it never needs to be known.

Question 10. 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.
(i) How many different quadrilaterals do they form if the quadrilateral is allowed to be self-intersecting or non-convex?
(ii) How many of these quadrilaterals are self-intersecting? How many are convex?
Answer:
(i) A 4-gon on these points is fixed by the order of its vertices, but each 4-gon has 8 names (4 starting points × 2 directions). There are 4! = 24 orders of 4 letters, so the number of different 4-gons is \( \frac{24}{8} \) = 3.
They are ABCD, ABDC and ACBD. (Equivalently, a 4-gon is fixed by which pair of segments are its diagonals, and the 6 segments joining the points form 3 such pairs: {AC, BD}, {AD, BC}, {AB, CD}.)
So the four points always form 3 different quadrilaterals.
(ii) There are two cases.

Case 1: no point lies inside the triangle of the other three (convex position). Exactly 1 is convex (the one going round the outside) and the other 2 are self-intersecting (their crossing sides are the diagonals of the convex one).
Case 2: one point lies inside the triangle of the other three. All 3 are non-convex, each with its dent at the inside point: 0 self-intersecting and 0 convex.
Summary: convex position: 1 convex, 0 non-convex, 2 self-intersecting; one point inside: 0 convex, 3 non-convex, 0 self-intersecting.

Question 11. 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 Fig. 12.39.
(i) If PB = 2 find QC. (Hint: See the dotted line)
*(ii) If PB = \( \frac{1}{3} \) find QC. (Hint: See the dotted lines)
*(iii) If PB = \( \frac{2}{3} \) find QC. (Hint: Divide both PB and AP suitably.)
*(iv) Show that if \( \frac{AP}{PB} \) is a rational number, then \( \frac{AP}{PB} \) = \( \frac{AQ}{QC} \). In Grade 10, you will prove this equality without assuming \( \frac{AP}{PB} \) is rational. That will require a new idea.
Answer:
Key fact (equal steps give equal steps). Let A = X₀, X₁, X₂, … be equally spaced on AB, and let the lines through them parallel to BC meet AC at A = Y₀, Y₁, Y₂, …. Then Y₀, Y₁, Y₂, … are also equally spaced.
Reason: in △AX₂Y₂, X₁ is the midpoint of AX₂ and X₁Y₁ ∥ X₂Y₂, so AY₁ = Y₁Y₂ (Theorem 7). Next, X₁X₃Y₃Y₁ is a trapezium with X₂ the midpoint of X₁X₃ and X₂Y₂ parallel to its parallel sides, so by Question 7(i) Y₂ is the midpoint of Y₁Y₃ and Y₁Y₂ = Y₂Y₃. Continuing, all steps on AC are equal.
(i) PB = 2. Mark the midpoint X of PB, so AP = PX = XB = 1: AB is 3 equal steps. PQ and the dotted line through X cut AC into 3 equal steps, the first being AQ = √5. So QC = 2 × √5 = 2√5.
(ii) PB = \( \frac{1}{3} \). Split AP into 3 steps of \( \frac{1}{3} \). Then AB is 4 equal steps, 3 in AP and 1 in PB. So AC is 4 equal steps with 3 in AQ: one step = \( \frac{\sqrt{5}}{3} \), and QC = \( \frac{\sqrt{5}}{3} \).
(iii) PB = \( \frac{2}{3} \). Use steps of \( \frac{1}{3} \): AP has 3 steps and PB has 2. AC is then 5 equal steps, 3 in AQ and 2 in QC: one step = \( \frac{\sqrt{5}}{3} \), QC = 2 × \( \frac{\sqrt{5}}{3} \) = \( \frac{2\sqrt{5}}{3} \).
(iv) Let \( \frac{AP}{PB} \) = \( \frac{m}{n} \) with m, n positive integers, and put d = \( \frac{AP}{m} \) = \( \frac{PB}{n} \).
Split AP into m steps of length d and PB into n steps of length d, so AB has m + n equal steps and P ends the m-th step. Draw lines parallel to BC through all the division points. By the key fact, AC is cut into m + n equal steps of some length e.
The line through P is PQ itself, so AQ = me and QC = ne. Therefore \( \frac{AQ}{QC} \) = \( \frac{me}{ne} \) = \( \frac{m}{n} \) = \( \frac{AP}{PB} \).
Check: (i) \( \frac{1}{2} \) = \( \frac{\sqrt{5}}{2\sqrt{5}} \); (ii) \( \frac{1}{1/3} \) = 3 = \( \frac{\sqrt{5}}{\sqrt{5}/3} \); (iii) \( \frac{1}{2/3} \) = \( \frac{3}{2} \) = \( \frac{\sqrt{5}}{2\sqrt{5}/3} \).

Teacher's Note:
1. Turn every part into 'count equal steps on AB, copy the count on AC'.
2. Part (iv) is the rational case of the Basic Proportionality Theorem of Grade 10.

Question 12. (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 Exercise 11 above and a third using the Centroid Theorem.

Answer:
(i) Let DM meet AC at X and BN meet AC at Y.
MBND is a parallelogram. MB = \( \frac{1}{2} \)AB = \( \frac{1}{2} \)DC = DN, and MB ∥ DN (they lie on AB and DC). By Theorem 5, MBND is a parallelogram, so DM ∥ NB.
1. In △ABY: M is the midpoint of AB and MX ∥ BY. By Theorem 7, X is the midpoint of AY: AX = XY.
2. In △CDX: N is the midpoint of CD and NY ∥ DX. By Theorem 7, Y is the midpoint of CX: CY = YX.
Hence AX = XY = YC, so DM and BN trisect AC.
(ii) Three ways to trisect PQ.
Method 1 (part (i)). Take any point R off line PQ. Complete the parallelogram PRQS with PQ as a diagonal (through P draw the parallel to RQ, through Q the parallel to RP; they meet at S). Mark the midpoints M of PR and N of QS. SM and RN cut PQ at its trisection points (part (i) with A = P, B = R, C = Q, D = S).
Method 2 (Exercise 11). Draw any ray from P not along PQ and mark three equal steps PU₁ = U₁U₂ = U₂U₃ with compasses. Join U₃Q and draw lines through U₁ and U₂ parallel to U₃Q. By the key fact of Exercise 11, these cut PQ into three equal parts.

Method 3 (Centroid Theorem). Take any point B off line PQ and extend BQ to C with QC = BQ. Then Q is the midpoint of BC and PQ is a median of △PBC. Let R be the midpoint of PC; the median BR meets PQ at the centroid G, with PG : GQ = 2 : 1, so GQ = \( \frac{1}{3} \)PQ. The midpoint H of PG gives PH = HG = \( \frac{1}{3} \)PQ. So H and G trisect PQ.

Teacher's Note:
1. The parallelogram MBND gives the parallel lines needed for Theorem 7 twice.
2. Method 2 is the usual textbook construction for dividing a segment.

Question 13. Is the Midpoint Theorem for Quadrilaterals (Theorem 9) 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.
(i) Can you explain why this is so? See the geometric reasoning in the text.
(ii) There is one exception: In a very special case the Varignon parallelogram becomes a single segment. When will this happen?
(iii) Does your reasoning apply even when ABCD is non-planar?
Answer:

(i) The proof in the text only uses the Midpoint Theorem in the four triangles ABC, ADC, BCD and BAD:
PQ ∥ AC, PQ = \( \frac{1}{2} \)AC (△ABC); SR ∥ AC, SR = \( \frac{1}{2} \)AC (△ADC);
QR ∥ BD, QR = \( \frac{1}{2} \)BD (△BCD); PS ∥ BD, PS = \( \frac{1}{2} \)BD (△BAD).
These four triangles exist whatever the shape of ABCD (convex, non-convex or self-intersecting), because each uses three of the four points and no three are collinear. So PQRS always has both pairs of opposite sides parallel and equal: it is a parallelogram.
(ii) The argument needs AC and BD to point in different directions. If AC ∥ BD, all four sides of PQRS are parallel to one direction, so P, Q, R, S are on one line and the 'parallelogram' collapses to a segment.
For an ordinary (convex or non-convex) quadrilateral the lines AC and BD always meet, so this can happen only for a self-intersecting ABCD whose 'diagonals' AC and BD are parallel.
Example: A(0, 0), B(0, 1), C(2, 0), D(2, 1). AC ∥ BD, and P = (0, \( \frac{1}{2} \)), Q = (1, \( \frac{1}{2} \)), R = (2, \( \frac{1}{2} \)), S = (1, \( \frac{1}{2} \)) all lie on one line (Q and S even coincide).
*(iii) Yes. Even if A, B, C, D are not in one plane, each of the triangles ABC, ADC, BCD, BAD lies in its own plane and the Midpoint Theorem holds in each. So PQ ∥ AC ∥ SR and PQ = SR = \( \frac{1}{2} \)AC.
Two lines parallel to the same line are parallel, and two distinct parallel lines lie in one plane; so P, Q, R, S are coplanar and PQRS has a pair of opposite sides equal and parallel. Hence PQRS is a planar parallelogram even for a non-planar ABCD.

Teacher's Note:
1. The proof never uses the shape of ABCD, only its four triangles.
2. The only failure is when both families of sides are parallel to one line.

Question 14. 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 Theorem 9.
Answer:
No. Counterexample: square ABCD with A(0, 0), B(4, 0), C(4, 4), D(0, 4), and P(1, 0) on AB, Q(4, 1) on BC, R(3, 4) on CD, S(0, 3) on DA.

Each corner triangle APS, BQP, CRQ, DSR is a right triangle with legs 1 and 3, so they are congruent (SAS) and PQ = QR = RS = SP.
So PQRS has opposite sides equal and is a parallelogram (Theorem 2), in fact a square. But AP = 1 ≠ 2, so P is not the midpoint of AB (similarly for Q, R, S).
Hence the converse of Theorem 9 is false.
(In any parallelogram ABCD, any two lines through the centre meeting the four sides give such a PQRS, because the four points are then symmetric about the centre.)

Question 15. (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 (Theorem 7). Start by extending PQ up to a suitable point S.

Answer:
(i) Given: P, Q are the midpoints of AB, AC in △ABC. To prove: PQ ∥ BC and PQ = \( \frac{1}{2} \)BC.
How far to extend. If △APQ ≅ △CSQ, then CS would be equal and parallel to AP, i.e. to PB, and Theorem 5 would give the parallelogram PBCS. We already have AQ = CQ and ∠AQP = ∠CQS (vertically opposite); SAS also needs QS = QP. So extend PQ beyond Q by its own length: QS = PQ.
1. In △APQ and △CSQ: AQ = CQ (Q is the midpoint of AC), PQ = SQ (construction), ∠AQP = ∠CQS (vertically opposite). So △APQ ≅ △CSQ (SAS).
2. Hence CS = AP and ∠QCS = ∠QAP.
3. ∠QCS and ∠QAP are alternate angles for lines AB and CS with transversal AC, so CS ∥ AB, i.e. CS ∥ PB.
4. Also CS = AP = PB.
5. In PBCS the opposite sides PB and CS are equal and parallel, so PBCS is a parallelogram (Theorem 5).
6. So PS ∥ BC and PS = BC. As Q is on PS with PQ = \( \frac{1}{2} \)PS: PQ ∥ BC and PQ = \( \frac{1}{2} \)BC.
(ii) Given: P is the midpoint of AB and the line through P parallel to BC meets AC at Q. To prove: Q is the midpoint of AC (and PQ = \( \frac{1}{2} \)BC).
Construction: extend PQ beyond Q to S so that PS = BC.
1. PS ∥ BC (given) and PS = BC (construction), so PBCS is a parallelogram (Theorem 5).
2. Hence CS ∥ BP, i.e. CS ∥ AB, and CS = BP = AP.
3. In △APQ and △CSQ: AP = CS (step 2); ∠PAQ = ∠SCQ (alternate angles, AB ∥ CS, transversal AC); ∠APQ = ∠CSQ (alternate angles, AB ∥ CS, transversal PS). So △APQ ≅ △CSQ (ASA).
4. Therefore AQ = CQ, so Q is the midpoint of AC. Also PQ = SQ, so PQ = \( \frac{1}{2} \)PS = \( \frac{1}{2} \)BC.

Teacher's Note:
1. In (i) the aim (a congruence) tells us where to put S.
2. In (ii) S is placed using length (PS = BC) instead, and the congruence comes at the end.

Question 16. * Review all the properties of a rhombus/rectangle/square that you proved in Grade 8. Formulate a converse of each. Decide if the converse is true. There are many possibilities here!
Answer:
Here '4-gon' means a planar, non-self-intersecting quadrilateral. Each line gives: property → converse → verdict.
1. Rhombus: all sides equal → a 4-gon with all sides equal is a rhombus → true (opposite sides equal gives a parallelogram, Theorem 2).
2. Rhombus: diagonals bisect each other at right angles → a 4-gon whose diagonals bisect each other at right angles is a rhombus → true (Ex. 12.2, Q3(ii)).
3. Rhombus: diagonals perpendicular → a 4-gon with perpendicular diagonals is a rhombus → false (a kite); true for a parallelogram.
4. Rhombus: each diagonal bisects the angles at its ends → a 4-gon whose diagonals bisect all its angles is a rhombus → true (Ex. 12.2, Q3(i)).
5. Rhombus: a diagonal bisects the angles at its ends → a parallelogram in which a diagonal bisects a vertex angle is a rhombus → true (Ex. 12.2, Q2); false for a general 4-gon (a kite).
6. Rectangle: all angles 90° → a 4-gon with all angles equal is a rectangle → true (each angle is 360° ÷ 4 = 90°).
7. Rectangle: diagonals equal → a 4-gon with equal diagonals is a rectangle → false (isosceles trapezium); true for a parallelogram (Ex. 12.2, Q1(iii)).
8. Rectangle: diagonals equal and bisect each other → a 4-gon whose diagonals are equal and bisect each other is a rectangle → true.
9. Square: all sides equal and all angles 90° → a 4-gon with all sides equal and one right angle is a square → true (a rhombus with a right angle).
10. Square: diagonals equal and perpendicular bisectors of each other → a 4-gon whose diagonals are equal and bisect each other at right angles is a square → true (a rhombus with equal diagonals, hence right angles).
11. Square: diagonals equal and perpendicular → a 4-gon whose diagonals are equal and perpendicular is a square → false (a kite with equal perpendicular diagonals); true for a parallelogram.
Proof of one 'true' entry (8): diagonals bisecting each other make the 4-gon a parallelogram (Theorem 4), and a parallelogram with equal diagonals is a rectangle (Ex. 12.2, Q1(iii)).

Question 17. * 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:
Set-up. Draw diagonal AC. The non-planar quadrilateral ABCD is made of △ABC and △ADC hinged along AC, lying in different planes.
Key fact. At A the rays AB, AC, AD are not in one plane, and then ∠BAD < ∠BAC + ∠CAD.
Reason (hinge picture): if △ADC is turned about AC until it lies flat in the plane of ABC on the side away from B, ∠BAD becomes exactly ∠BAC + ∠CAD. As △ADC is turned out of the plane, D moves closer to B, so ∠BAD becomes smaller. Similarly at C: ∠BCD < ∠BCA + ∠ACD.
The angles at B and D are just the angles of the triangles: ∠B = ∠ABC, ∠D = ∠ADC.
Adding: ∠A + ∠B + ∠C + ∠D < (∠BAC + ∠ABC + ∠BCA) + (∠CAD + ∠ADC + ∠ACD) = 180° + 180° = 360°.
So the angle sum of a non-planar quadrilateral is always less than 360°.
During the rotation ∠B and ∠D stay fixed (rigid triangles), while ∠A and ∠C steadily decrease as △ADC is turned from the flat position towards folding onto △ABC. When the triangles are folded together, ∠A and ∠C become 0.
Sum equal to 2°: yes. Choose thin triangles with ∠ABC = ∠ADC = 0.9° (△ADC a mirror image of △ABC). Flat, the sum is 360°; folded completely (D falling on B), it is 0° + 0.9° + 0° + 0.9° = 1.8°.
The sum changes continuously from 360° to 1.8° during the rotation, so at some in-between (non-planar) position it is exactly 2°.

Teacher's Note:
1. The hinge picture shows why angles at A and C shrink while B and D do not.
2. The '2°' part uses a continuity argument: the sum passes through every value between 1.8° and 360°.

Question 18. 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 Fig. 12.42 so that the diagonals EO and E′O′ are collinear with O = E′. 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 EO′ until EO matches E′O′. Verify that your cutout exactly matches S′O′M′E′.
(i) Prove the exact match of SOME with S′O′M′E′ using four parallelograms.
(ii) Follow the described procedure along each diagonal of the starting coloured 4-gon in both directions to get 4 new 4-gons. Using these 4 copies, repeat the procedure 4 more times to place 4 new copies, resulting in a 3 by 3 grid of 9 copies exactly like the 9 coloured copies in Fig. 12.42. These 9 copies meet each other only at vertices and once again the blank spaces among them also form 4-gons congruent to SOME!
Answer:
(i) When the cutout slides so that E moves to E′ (= O), every point moves the same distance EO in the same direction. So, with S′, O′, M′ the new positions of S, O, M: SS′, OO′, MM′ are all parallel to EE′ and SS′ = OO′ = MM′ = EE′.
This gives four parallelograms (one pair of opposite sides equal and parallel, Theorem 5):
1. ESS′E′: EE′ ∥ SS′ and EE′ = SS′, so E′S′ = ES and E′S′ ∥ ES.
2. SOO′S′: SS′ ∥ OO′ and SS′ = OO′, so S′O′ = SO and S′O′ ∥ SO.
3. OMM′O′: OO′ ∥ MM′ and OO′ = MM′, so O′M′ = OM and O′M′ ∥ OM.
4. EMM′E′: EE′ ∥ MM′ and EE′ = MM′, so E′M′ = EM and E′M′ ∥ EM.
So every side of S′O′M′E′ is equal and parallel to the matching side of SOME, and E′O′ = EO. Hence △E′S′O′ ≅ △ESO and △E′M′O′ ≅ △EMO (SSS), and the slid cutout matches S′O′M′E′ exactly.
(ii) Sliding SOME forwards and backwards along EO, and along SM, gives 4 new copies, each touching the first only at a vertex. Sliding these again gives 4 more, completing the 3 × 3 grid of 9 copies (green in the figure).
Why the gaps are copies of SOME. Each gap (yellow) is surrounded by four copies: SOME, SOME slid along SM, SOME slid along EO, and SOME slid along both. Let R be the half-turn of SOME about the midpoint of OM (as in Method 1).
1. R shares the whole side OM with SOME.
2. A half-turn of R about the midpoint of its side through M gives SOME slid along SM (two half-turns about midpoints of adjacent sides make a slide along a diagonal, In-text Question 26). So R shares that full side with the slid copy.
3. In the same way R shares full sides with SOME slid along EO and with SOME slid along both diagonals.
So R fits exactly into the gap: each gap is a copy of SOME turned through 180°. At each corner of a gap two slid copies and two gaps meet, supplying the four different angles of SOME, which total 360°. Continuing the slides and filling gaps tiles the plane, giving the same tiling as Methods 1 and 2.

Teacher's Note:
1. A slide moves every point by the same amount in the same direction; that is where the four parallelograms come from.
2. Link the gaps to Method 1: they are exactly the half-turned copies.

Question 19. Is there a 4-gon with given side lengths?
(i) Recall the following fact about triangles and check it by construction. For given positive numbers a, b, c, is there a triangle whose sides have these lengths? The answer is Yes exactly when the sum of any two numbers is greater than the third. If we arrange the numbers in increasing order (suppose a ≤ b ≤ c), then this amounts to requiring a + b > c. (Hint: Start by drawing a segment of length c.)
(ii) Suppose a 4-gon has 2, 5, 11 as three side lengths. Can the length of the fourth side be 100? Can it be 10? Can it be 1? What are the possible lengths of the fourth side?
(iii) For given positive numbers a, b, c, d, how will you decide if there is a 4-gon whose sides have these lengths?
Answer:
(i) Construction. Draw XY of length c. Draw an arc of radius b about X and an arc of radius a about Y.

1. If a + b > c, the arcs meet at a point Z and △XYZ has sides c, b, a.
2. If a + b = c, the arcs just touch at a point on XY: only a flat 'triangle' (a segment).
3. If a + b < c, the arcs are too short to meet: no triangle.
When a ≤ b ≤ c, the other two conditions a + c > b and b + c > a hold automatically, so a + b > c is the only check. For example 3, 4, 6 works (3 + 4 = 7 > 6) but 2, 3, 6 does not (2 + 3 = 5 < 6).
(ii) Going round three sides from one end of the fourth side to the other is longer than the straight side (triangle inequality twice, using a diagonal). So each side must be less than the sum of the other three. Let the fourth side be x.
1. x = 100: need 100 < 2 + 5 + 11 = 18. False, so not possible.
2. x = 10: the largest side is 11 and 11 < 2 + 5 + 10 = 17; also 10 < 18. Possible.
3. x = 1: need 11 < 2 + 5 + 1 = 8. False, so not possible.
In general we need x < 18 and 11 < 7 + x, i.e. x > 4 (the conditions for sides 2 and 5 then hold automatically). So 4 < x < 18.
(iii) Rule: 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 (equivalently, each is less than the sum of the other three).

Necessary: if ABCD has AB = a, BC = b, CD = c, DA = d, then using diagonal AC and the triangle inequality twice, d = DA < DC + CA < c + (a + b). The same works for each side.
Sufficient: let a ≤ b ≤ c ≤ d with d < a + b + c. We want a diagonal length x for which both triangles (a, b, x) and (c, d, x) exist, i.e. b − a < x < a + b and d − c < x < c + d.
Such an x exists because each lower bound is below each upper bound: b − a < a + b, b − a < c + d, d − c < c + d, and d − c < a + b (this last one is exactly d < a + b + c).
Draw AC of length x, build △ABC with AB = a, BC = b on one side and △ADC with CD = c, DA = d on the other. ABCD is the required 4-gon.

Teacher's Note:
1. For three given sides, write both conditions: x below the sum of the three, and the largest side below the sum of the rest.
2. Point out that the answer is a range (4 < x < 18), not a single value.

Question 20. Counting diagonals of a polygon.
(i) How should we define a diagonal of an n-gon? How many diagonals does an n-gon have? Make a table for small values of n. A 3-gon has no diagonals. A 4-gon has 2. How many diagonals does a 5-gon have? A 6-gon? Try to find a pattern and guess the answers for n = 7 and n = 8. Check your guesses by systematic counting.
(ii) Can you guess a formula for the number of diagonals? How many diagonals get added when we increase the number of sides by 1? Can you now justify why the formula you guessed is true for all n?
Answer:
(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).

Systematic counting: from each vertex there are diagonals to every vertex except itself and its two neighbours, i.e. to n − 3 vertices. Doing this at all n vertices gives n(n − 3), but each diagonal is counted twice (once from each end).
n = 3: 0; n = 4: 2; n = 5: 5; n = 6: 9; n = 7: 14; n = 8: 20.
Increase from the previous value: 2, 3, 4, 5, 6.
The pattern of increases suggests 9 + 5 = 14 for n = 7 and 14 + 6 = 20 for n = 8, and counting confirms it: \( \frac{7 \times 4}{2} \) = 14 and \( \frac{8 \times 5}{2} \) = 20.
(ii) Formula: D(n) = \( \frac{n(n-3)}{2} \).
Diagonals added when n goes up by 1. Insert a new vertex V between neighbouring vertices X and Y (side XY is replaced by XV and VY).
1. V is joined by diagonals to all old vertices except X and Y: n − 2 new diagonals.
2. The old side XY is no longer a side; it becomes a diagonal: 1 more.
So (n − 2) + 1 = n − 1 diagonals are added: D(n + 1) = D(n) + (n − 1).
Justification for all n. D(3) = 0 = \( \frac{3 \times 0}{2} \). If D(n) = \( \frac{n(n-3)}{2} \), then D(n + 1) = \( \frac{n(n-3)}{2} \) + (n − 1) = \( \frac{n^2 - 3n + 2n - 2}{2} \) = \( \frac{n^2 - n - 2}{2} \) = \( \frac{(n+1)(n-2)}{2} \),
which is the formula with n replaced by n + 1. Starting at n = 3 and adding one side at a time, D(n) = \( \frac{n(n-3)}{2} \) holds for every n. (The direct count in (i) also proves it.)

Question 21. 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:
Values. Diagonals from one vertex cut a pentagon into 3 triangles, a hexagon into 4, and so on:

n = 3: 1 triangle, 180°; n = 4: 2, 360°; n = 5: 3, 540°; n = 6: 4, 720°; n = 7: 5, 900°; n = 8: 6, 1080°.
Formula: sum of angles of an n-gon = (n − 2) × 180°.
Proof for a convex n-gon. From vertex A₁ draw the diagonals A₁A₃, A₁A₄, …, A₁A₍ₙ₋₁₎. They cut the polygon into the n − 2 triangles A₁A₂A₃, A₁A₃A₄, …, A₁A₍ₙ₋₁₎Aₙ, whose angles together make up exactly the angles of the polygon. So the sum is (n − 2) × 180°.
Proof for any (possibly non-convex) n-gon. Every polygon with n ≥ 4 has a diagonal lying inside it. (Take a vertex V with internal angle below 180°, such as the lowest vertex. Either the segment joining its two neighbours lies inside, or some vertex lies inside that triangle, and the one nearest to V can be joined to V.)
This diagonal cuts the n-gon into two polygons with k and n − k + 2 sides, both fewer than n, whose angles together make up the angles of the n-gon. If the formula is known for fewer sides, the sum is (k − 2) × 180° + (n − k + 2 − 2) × 180° = (n − 2) × 180°.
Starting from triangles (180°), the formula holds for all n.

Teacher's Note:
1. n − 2 triangles, not n: the vertex we fan from has no 'own' triangle on each side.
2. The non-convex proof builds up from smaller polygons, one diagonal at a time.

Question 22. Multiple converses to a theorem. Let us see how multiple statements can be considered converses to the Midpoint Theorem and how Theorem 7 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 = \( \frac{BC}{2} \).
Let us use these four named conditions to discuss various possible statements.
(i) A natural converse of the Midpoint Theorem would be: "If (PRLL) and (HALF) are true, then (P MID) and (Q MID) are true." Write out this statement fully. Experiment and see that it seems to be true! Prove the statement. (Hint: Try to run a proof of the Midpoint Theorem backwards.)
(ii) To see Theorem 7 as a converse of the Midpoint Theorem, we first write the Midpoint Theorem as follows: "Suppose P is the midpoint of side AB of △ABC and Q is a point on side AC. If Q is the midpoint of AC then PQ ∥ BC." (Also PQ = \( \frac{BC}{2} \), but let us set that aside for now.) Write the converse of the second sentence in quotes by keeping the first sentence the same. Verify that Theorem 7 is what you get! (And PQ = \( \frac{BC}{2} \) also follows.) Now write Theorem 7 in terms of the four named conditions.
(iii) The discussion so far suggests that it is reasonable to take two of the four listed statements as assumptions and ask if the other two are true. Verify that only one such combination remains to be examined. "If (P MID) and (HALF) are true, then can we conclude (Q MID) and/or (PRLL)?" Write this out in words. This is a precise version of question (1) stated before Theorem 7. Can you answer it?
In the text, why did we focus only on Theorem 7 among the multiple possibilities? Because Theorem 7 turns out to be particularly useful. It is still important to pursue various avenues while exploring a subject: often it is only after examining multiple possibilities that we can judge what is most valuable.
Answer:
(i) Statement: in △ABC let P be a point on side AB and Q a point on side AC. If PQ ∥ BC and PQ = \( \frac{BC}{2} \), then P is the midpoint of AB and Q is the midpoint of AC.
Proof. Let M be the midpoint of AB. The line through M parallel to BC meets AC at its midpoint N (Theorem 7), and MN ∥ BC, MN = \( \frac{BC}{2} \) (Midpoint Theorem).
Suppose P ≠ M. PQ and MN both run from line AB to line AC, are parallel to BC and have the same length \( \frac{BC}{2} \). So in PQNM the opposite sides PQ and MN are equal and parallel, and PQNM is a parallelogram (Theorem 5). Then PM ∥ QN.
But PM lies on line AB and QN on line AC, and these lines meet at A, so they are not parallel. This contradiction shows P = M: P is the midpoint of AB. Then PQ is the line through the midpoint of AB parallel to BC, so by Theorem 7, Q is the midpoint of AC. ◼
(ii) Keeping the first sentence, the converse of the second is: "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, and as shown in the text PQ = \( \frac{BC}{2} \) also follows.
In terms of the named conditions, Theorem 7 says: if (P MID) and (PRLL), then (Q MID) and (HALF).
(iii) There are 6 ways to choose two of the four conditions as assumptions:
1. (P MID) and (Q MID): the Midpoint Theorem (Theorem 6).
2. (PRLL) and (HALF): part (i), true.
3. (P MID) and (PRLL): Theorem 7, true.
4. (Q MID) and (PRLL): Theorem 7 with AB and AC swapped, true.
5. (Q MID) and (HALF): the same as (P MID) and (HALF) with AB and AC swapped.
6. (P MID) and (HALF): still to be examined.
So, up to swapping the sides, only one combination remains. In words: "In △ABC, P is the midpoint of AB and Q is a point on AC with PQ = \( \frac{BC}{2} \). Must Q be the midpoint of AC? Must PQ be parallel to BC?"
Answer: not necessarily. Let N be the midpoint of AC; then PN = \( \frac{BC}{2} \). The circle with centre P and radius \( \frac{BC}{2} \) passes through N and usually cuts line AC at a second point Q (the mirror image of N in the foot F of the perpendicular from P to AC). If this Q lies on side AC, then (P MID) and (HALF) hold, but Q ≠ N, so Q is not the midpoint and PQ is not parallel to BC (the only segment from P parallel to BC is PN).

Example: AB = 4 cm, AC = 5 cm, ∠A = 30°. The circle with centre P through N meets AC again at Q about 0.96 cm from A, which is on side AC. So (P MID) and (HALF) do not imply (Q MID) or (PRLL).
They do imply them in the special cases when the second point falls outside side AC, or coincides with N (when PN ⊥ AC); then Q must be N.
(For question (1) before Theorem 7, a segment PQ with P on AB, Q on AC, parallel to the third side and half its length: by part (i), P and Q must be the midpoints.)

Teacher's Note:
1. Name the conditions (P MID, Q MID, PRLL, HALF) to keep the six combinations straight.
2. The circle picture shows why a length condition alone cannot fix the position of Q.

Question 23. 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 must be proved: (a) the copies never overlap, (b) there are no gaps, and (c) the process can go on for ever so that every point of the plane is eventually covered.
Step 1: all three methods give the same arrangement. Let SOME be the 4-gon and R its half-turn about the midpoint of OM.
1. Method 3 places slid copies of SOME (slides along the diagonals SM and EO and their combinations) and fills each gap with R slid in the same way (End-of-Chapter Q18).
2. Method 1 uses only half-turns about midpoints of sides. Two half-turns about midpoints of adjacent sides make a slide along a diagonal (In-text Question 26), so Method 1 produces exactly the slid copies of SOME and of R.
3. Method 2 puts copies so that their Varignon parallelograms fill the chessboard grid. The Varignon parallelogram has sides \( \frac{1}{2} \)SM and \( \frac{1}{2} \)EO parallel to the diagonals, so moving two cells along the grid is again a slide along a diagonal: the same arrangement.
So it is enough to prove that this one arrangement is a tiling.
Step 2: a hexagon made of two copies. Put SOME and R together along OM. They form a hexagon H with vertices S, O, E*, S*, M, E (S*, E* are the images of S, E). The half-turn about the midpoint of OM maps H onto itself, so opposite sides of H are equal and parallel: SO and S*M, OE* and ME, E*S* and ES.

Step 3: hexagons fit in strips. Slide H along the diagonal from E to O. Side ME of H is equal and parallel to the opposite side E*O, and the slide carries M to E* and E to O; so side ME of the slid hexagon lands exactly on side E*O of H. The two hexagons share a full side and lie on opposite sides of it. Repeating forwards and backwards gives an endless strip.
Step 4: strips fit together. The upper edge of a strip is a zigzag of the sides SO and E*S* repeated; the lower edge is made of the opposite sides S*M and ES, which are equal and parallel to them. Sliding the whole strip along the diagonal SM carries its lower zigzag exactly onto the upper zigzag of the original strip, with no gap or overlap.
Step 5: the whole plane is covered. Each strip has a fixed width and the strips are stacked by the same slide in both directions, like rows of a brick wall. So they cover the plane, and since every hexagon is SOME + R, the plane is tiled by copies of SOME.
Check at vertices: at every vertex the four angles of SOME meet once each (∠1 + ∠2 + ∠3 + ∠4 = 360°), confirming there is no gap or overlap around any vertex. Hence each of the three methods gives a tiling.

Teacher's Note:
1. First reduce the three methods to one arrangement, then prove that one arrangement works.
2. The hexagon SOME + R has opposite sides equal and parallel, which is what makes strips stack.

Question 24. What fraction of the square is shaded?

Answer:
In Fig. 12.43 each side of the square is bisected, and each of the four lines joins a corner to the midpoint of a side, giving two pairs of parallel lines. Let the square ABCD have side 2 units, so its area is 2 × 2 = 4 square units.
Side of the shaded square. Take the parallel lines DL and KB (L the midpoint of AB, K the midpoint of DC). They bound the strip DKBL.
1. DK = LB = 1 and DK ∥ LB, so DKBL is a parallelogram (Theorem 5).
2. With base DK = 1 and height AD = 2, ar(DKBL) = 1 × 2 = 2.
3. DL is the hypotenuse of right △DAL with legs 2 and 1: DL = \( \sqrt{2^2 + 1^2} \) = √5.
4. With DL as base, the height of the parallelogram is the distance between the parallel lines DL and KB, which is exactly the side s of the shaded square (the other two lines are perpendicular to these). So √5 × s = 2, i.e. s = \( \frac{2}{\sqrt{5}} \).
Area of the shaded square: s² = \( \left(\frac{2}{\sqrt{5}}\right)^2 \) = \( \frac{4}{5} \).
Fraction: \( \frac{4/5}{4} \) = \( \frac{1}{5} \). The shaded square is \( \frac{1}{5} \) of the whole square.
(Check with the Midpoint Theorem: each of the four lines is cut by the others into three pieces in the ratio 2 : 2 : 1, of lengths \( \frac{2}{\sqrt{5}} \), \( \frac{2}{\sqrt{5}} \), \( \frac{1}{\sqrt{5}} \), adding to √5; the middle piece is a side of the shaded square.)

Teacher's Note:
1. Computing one area in two ways (base DK, then base DL) gives the side of the small square.
2. A classic extension: the pieces outside the shaded square can be rearranged into four more squares of the same size.