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ABCD has AB = CD = 6 and AB ∥ CD. Is it a parallelogram? Why?
Yes: one pair of opposite sides is equal and parallel (Theorem 5).
Answer: parallelogram
In quadrilateral ABCD, ∠A = ∠C = 70° and ∠B = ∠D. Find ∠B. Is ABCD a parallelogram?
360° − 140° = 220°, so ∠B = ∠D = 110°.
Opposite angles equal ⇒ parallelogram (∠A + ∠B = 180°).
Answer: 110°, yes
The diagonals AC and BD of ABCD meet at O with OA = 5, OC = 5, OB = 4, OD = 4. Name the figure.
The diagonals bisect each other ⇒ parallelogram.
Answer: parallelogram
In parallelogram ABCD, points P, Q on BD satisfy BP = DQ. Show APCQ is a parallelogram.
Diagonals AC and BD bisect at O; OP = OB − BP = OD − DQ = OQ.
So the diagonals AC and PQ of APCQ bisect each other.
Answer: proved
A parallelogram’s diagonal bisects ∠A. Show it is a rhombus.
∠BAC = ∠DAC = ∠ACB (alternate angles as AD ∥ BC).
So BA = BC: adjacent sides equal ⇒ rhombus.
Answer: rhombus
In ΔABC, D, E are midpoints of AB, AC and BC = 14 cm. Find DE.
DE = BC ÷ 2 = 7 cm and DE ∥ BC.
Answer: 7 cm
A triangle has sides 10, 12, 14 cm. What is the perimeter of the triangle joining the midpoints?
Each side is half: 5, 6, 7. Perimeter 18 cm (half of 36).
Answer: 18 cm
In ΔABC, P is the midpoint of AB, PQ ∥ BC meets AC at Q, and AC = 9. Find AQ.
Theorem 7: Q is the midpoint of AC, so AQ = 4.5.
Answer: 4.5
The median AD of ΔABC is 12 cm and G is the centroid. Find AG and GD.
AG : GD = 2 : 1, so AG = 8 cm and GD = 4 cm.
Answer: 8 cm, 4 cm
In trapezium ABCD with AB ∥ CD, AB = 10, CD = 6, E and F are midpoints of AD and BC. Find EF.
EF ∥ AB and EF = (10 + 6) ÷ 2 = 8.
Answer: 8
ABCD has diagonals AC = 10 and BD = 6. What kind of parallelogram is PQRS and what are its sides?
PQ = SR = AC ÷ 2 = 5; QR = PS = BD ÷ 2 = 3. A parallelogram with sides 5 and 3 (a rectangle if AC ⟂ BD).
Answer: sides 5 and 3
In a non-convex quadrilateral three angles are 40°, 50° and 70°. Find the fourth. Is it more than 180°?
360° − 160° = 200°, more than 180°: it is the dent.
Answer: 200°
How many diagonals does a hexagon have?
n(n − 3) ÷ 2 = 6 × 3 ÷ 2 = 9.
Answer: 9
What is the sum of the angles of a convex octagon?
(8 − 2) × 180° = 1080°.
Answer: 1080°
Why can a regular pentagon not tile the plane?
Each angle is 108°. 3 × 108° = 324° leaves a gap and 4 × 108° = 432° overlaps; 360° is not a multiple of 108°.
Answer: 108° does not divide 360°
How can any triangle tile the plane?
Two copies (one rotated 180° about a side’s midpoint) form a parallelogram, and parallelograms tile the plane.
Answer: via a parallelogram
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NOPE and SILY have three collinear vertices (a triangle or overlapping sides), BENT is non-planar, CUTS is self-intersecting, OPENS is not closed correctly (more than 4 sides/points on a side). DART is a (non-convex) quadrilateral. This is why Definition 1 requires four distinct points, no three collinear, in one plane, with no other point on more than one side.
Then each pair of adjacent angles adds up to 180° (360° ÷ 2). Conversely supplementary adjacent angles give equal opposite angles. Theorem 3: “If each pair of adjacent angles in a quadrilateral ABCD are supplementary, then ABCD is a parallelogram.”
ΔACD ≅ ΔCAB (SSS: AD = CB, DC = AB, AC common). So ∠DAC = ∠BCA and ∠DCA = ∠BAC. Alternate angles equal give AD ∥ BC and AB ∥ DC.
360° = ∠A + ∠B + ∠C + ∠D = 2(∠A + ∠B) so ∠A + ∠B = 180°, giving BC ∥ AD (interior angles on the same side of AB). Likewise ∠B + ∠C = 180° gives AB ∥ DC.
With EA = EC, EB = ED and ∠AED = ∠CEB (vertically opposite), ΔAED ≅ ΔCEB (SAS). So ∠DAE = ∠BCE, and AD ∥ BC. Similarly ΔEAB ≅ ΔECD gives AB ∥ DC.
AB ∥ DC and AB = DC. The diagonals meet at E (see Ex 12.1 Q3). ΔEAB ≅ ΔECD (ASA: ∠EAB = ∠ECD, AB = CD, ∠EBA = ∠EDC). So EA = EC and EB = ED. By Theorem 4 ABCD is a parallelogram.
Too many unknowns appear at once, so we study one midpoint segment first and use the Midpoint Theorem; then all three sides give four congruent triangles.
See the figure in the notes: l ∥ BA through C meets PQ at R. ΔAPQ ≅ ΔCRQ (AAS) ⇒ CR = AP = BP and PQ = QR. BCRP has BP = CR and BP ∥ CR, so it is a parallelogram: PQ ∥ BC and PQ = ½ PR = ½ BC.
Draw the line through P parallel to BC meeting AC at Q; draw l through C parallel to BA meeting PQ at R. BCRP is a parallelogram so CR = BP = AP; ΔAPQ ≅ ΔCRQ (AAS) ⇒ AQ = QC. Alternatively let M be the midpoint of AC; PM ∥ BC by Theorem 6, and by the uniqueness of the parallel through P, line PM = line PQ, so Q = M. A theorem is used to prove its converse because of the uniqueness of parallels.
Take medians CP and BQ meeting at M. X, Y are midpoints of BM and CM. PQ ∥ BC, PQ = BC/2 (triangle ABC) and XY ∥ BC, XY = BC/2 (triangle MBC). ΔMPQ ≅ ΔMYX (ASA), so MP = MY = YC and MQ = MX = XB. So CM : MP = 2 : 1 = BM : MQ. The same with medians AR and BQ gives N with BN : NQ = 2 : 1, so N = M. Alternative: extend AM to S with BSCM a parallelogram.
In ΔABC and ΔADC: PQ ∥ AC ∥ SR; in ΔBCD and ΔBAD: QR ∥ BD ∥ PS. So PQRS has both pairs of opposite sides parallel.
(i) Adjacent to AB: AD and BC; opposite to AB: CD. (ii) Adjacent to ∠A: ∠B and ∠D; opposite: ∠C. (iii) A pair of opposite sides: two sides with no common endpoint. A pair of opposite angles: two angles whose vertices are not adjacent (they share no side).
At vertex A let ∠BAD (less than 180°) be the angle between rays AB and AD. If the opposite vertex C lies inside this angle, the internal angle at A is ∠BAD. Otherwise (the dent of a non-convex quadrilateral) it is 360° − ∠BAD, which is more than 180°.
AB ∥ DC: no, it must be convex (a non-self-intersecting quadrilateral with a pair of parallel sides is convex). AB = CD: yes it can be non-convex, e.g. A(0, 0), B(5, 0), C(8, 4), D(4, 1) (AB = CD = 5, D is a dent). ∠A = ∠C: yes, e.g. the symmetric dart A(0, 2), B(5, 0), C(0, −2), D(1, 0).
The 7 regions: inside ΔABC: non-convex (D is the dent). Beyond AC (across AC from B, inside the angle at B): convex. Beyond AB or beyond BC (across one side only): self-intersecting. The three regions at the vertices (vertically opposite the triangle’s angles): non-convex.
No. If sides AB and CD intersect, the two lines meet in a point and lie in one plane, so all four points A, B, C, D lie in that plane.
(i) True. (ii) False: every rhombus has perpendicular diagonals, so it need not be a square (a square also needs equal diagonals). (iii) True.
∠BAC = ∠DAC. AD ∥ BC gives ∠DAC = ∠ACB, and AB ∥ DC gives ∠BAC = ∠ACD. So ∠ACB = ∠ACD: AC bisects ∠C. Also ∠BAC = ∠ACB means BA = BC; adjacent sides of the parallelogram are equal, so it is a rhombus.
(i) Yes: ΔABC ≅ ΔADC (ASA) and ΔABD ≅ ΔCBD (ASA) give AB = AD = CB = CD. (ii) Yes: it is a parallelogram with perpendicular diagonals, so by SAS all sides are equal. (iii) No (isosceles trapezium). Add: the diagonals also bisect each other (ABCD is a parallelogram).
Adjacent angles A and B sum to 180°, so half of each sum to 90°: in the triangle formed by AB and the two bisectors, the third angle is 90°. The same holds at each of the four meeting points, so PQRS has four right angles: a rectangle. If AB = BC, ABCD is a rhombus, the bisectors are the diagonals and meet in one point, so no rectangle forms.
(i) PQ = BC/2 = BR = RC, PR = AC/2 = AQ = QC, QR = AB/2 = AP = PB. So ΔPQR ≅ ΔQPA (SSS), and ΔPQR ≅ ΔRBP and ΔQRC. (ii) Yes: through each vertex of PQR draw the line parallel to the opposite side; the three lines form ΔABC, with P, Q, R as midpoints.
MN ∥ BC and M is the midpoint of AB. In ΔABD, the line through M parallel to BD meets AD at its midpoint (Theorem 7). MN passes through that point, so MN bisects AD.
Let E be the midpoint of AD. In ΔADC, EG ∥ DC, EG = DC/2; in ΔDAB, EH ∥ AB, EH = AB/2. Both lines through E are parallel to AB, so E, G, H are collinear and GH ∥ AB, with GH = |AB − CD| ÷ 2. If AB = CD then G = H (ABCD is a parallelogram), so there is no segment GH.
(i) PQRS is a parallelogram, so PR and QS bisect each other. (ii) PQ = AC/2, QR = BD/2. If AC = BD then PQ = QR: PQRS is a rhombus, so PR ⟂ QS. Converse: if PR ⟂ QS then PQRS is a rhombus, PQ = QR, so AC = BD. It is true.
(i) The sides give A + B = 2P, B + C = 2Q, C + D = 2R, D + A = 2S. Choose a position for A′, reflect it in P to get B′, B′ in Q to get C′, C′ in R to get D′; D′ reflected in S returns to A′ (because PQRS is a parallelogram). If A′ is placed as A was relative to PQRS, A′B′C′D′ is a congruent copy of ABCD; the half-turn about the centre of PQRS gives another copy. (ii) Reflections in P, Q, R, S give ΔSDR ≅ ΔSD′R etc. and show S lies on A′D′ with SA′ = SD′. (iii) PQRS is a square ⇔ PQ = QR and PQ ⟂ QR ⇔ AC = BD and AC ⟂ BD (since PQ = AC/2 ∥ AC, QR = BD/2 ∥ BD).
Its angle is 108°. Angles around a vertex must total 360°. 3 × 108° = 324° leaves a gap and 4 × 108° = 432° overlaps; 360 ÷ 108 is not an integer.
Method 1: rotate a copy by 180° about the midpoint of each side and repeat. Method 2: draw the Varignon grid of the DART and place copies in the pattern of Fig. 12.31. Both work because the four angles add up to 360° and one of each fits around a vertex. (Method 1 is quicker to describe.)
Rotate a copy of the triangle by 180° about the midpoint of one side: the two triangles form a parallelogram, which tiles the plane (sheared grid).
Parallel, equally spaced lines cut equal pieces on any transversal. If the line’s endpoints lie on lines 2k spaces apart, the midpoint is where the segment meets the line k spaces from each end. Justification: through the midpoint of one side of a trapezium a parallel to the bases bisects the other side (Ex 7(i) / Theorem 7), applied repeatedly.
In ΔAED: ∠A + ∠D + ∠AED = 180°; in ΔBEC: ∠B + ∠C + ∠BEC = 180°; ∠AED = ∠BEC (vertically opposite). Adding: ∠A + ∠B + ∠C + ∠D = 360° − 2∠AED < 360°. For 2° we need ∠AED = 179°, which is possible with a long thin figure.
The diagonals AC and BD bisect each other at O. OP = OD − DP = OB − BQ = OQ. So in APCQ the diagonals AC and PQ bisect each other at O: a parallelogram (Theorem 4).
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, so the crease ∥ BC. By Theorem 7, the line through the midpoint M of AB parallel to BC bisects AC. So the crease marks the midpoints of AB and AC.
Medians meet at M with MA = ⅔ AP etc. Placing ΔMPB so PB lies on PC (B to C) the two triangles at P form a triangle with M, P, M′ collinear (the angles at P add to 180°): sides MM′ = 2MP = ⅓·2·AP, MC = ⅔ CR and M′C = MB = ⅔ BQ. So the sides are ⅔ of the three medians, and the same for the other two pairs: three congruent triangles.
The medians of a triangle with sides m₁, m₂, m₃ (themselves medians of ABC scaled by ⅔) are ¾ of the sides of ABC scaled by ⅔, that is ½ of AB, BC, AC. The new triangles have sides ⅔ of these medians: AB/3, BC/3, AC/3.
Cut along the median AP (P the midpoint of BC) and then cut ΔAPB along its median PR (R the midpoint of AB). Rotate ΔBPR by 180° about R: B goes to A and P goes to P′. ΔARP and ΔARP′ form ΔAPP′ with sides AP, AP′ = PB = BC/2 and PP′ = 2PR = AC (Midpoint Theorem). ΔAPC has sides AP, PC = BC/2, AC. By SSS ΔAPP′ ≅ ΔAPC.
ΔOPB ≅ ΔOQD: OB = OD, ∠POB = ∠QOD (vertically opposite), ∠OBP = ∠ODQ (alternate angles, AB ∥ CD): ASA. So OP = OQ.
EF = (3 + 5) ÷ 2 = 4 cm, and each part has height h/2. Area AEFD = (3 + 4)/2 × h/2 = 7h/4; area EBCF = (4 + 5)/2 × h/2 = 9h/4. Ratio 7 : 9.
(i) Three different quadrilaterals can be formed (pairing the points into three different cyclic orders). (ii) If the points are in convex position: one convex, two self-intersecting. If one point lies inside the triangle of the other three: all three are non-convex and none is self-intersecting or convex.
(i) PB = 2: cut AB into three equal parts of 1 (parallels bisect/trisect AC): QC = 10. (ii) PB = 1/3: AP : PB = 3 : 1, so QC = 5/3. (iii) PB = 2/3: AP : PB = 3 : 2, QC = 10/3. (iv) In all cases AP/PB = AQ/QC.
AM ∥ NC and AM = NC (halves of equal parallel sides), so DM ∥ NB (DMBN is a parallelogram). Let DM meet AC at X and BN meet AC at Y. In ΔABY, M is the midpoint of AB and MX ∥ BY, so X is the midpoint of AY. In ΔCDX, N is the midpoint of CD and NY ∥ DX, so Y is the midpoint of CX. Hence AX = XY = YC. To trisect PQ: draw a parallelogram with diagonal PQ, mark midpoints M, N of two opposite sides, join; or draw a ray from P, mark three equal steps, join the third mark to Q and draw parallels (Ex 11); or build a triangle with PQ as a median and use the centroid (2 : 1) then bisect the longer part.
It remains true. (i) The proof only uses the Midpoint Theorem in ΔABC, ΔADC, ΔBCD, ΔBAD, which holds for any four points. (ii) When AC ∥ BD (the diagonals’ lines parallel), PQ, QR, … all lie along one line, so PQRS degenerates to a segment. (iii) Yes, each triangle lies in a plane, so the same reasoning gives PQ ∥ AC ∥ SR and QR ∥ BD ∥ PS, and PQRS is a (planar) parallelogram.
No. In a parallelogram ABCD take AP = CR = x and BQ = DS = y: PQRS is a parallelogram for any x, y.
(i) Extend PQ beyond Q to S with QS = PQ. Then ΔAPQ ≅ ΔCSQ (SAS, vertical angles), so CS = AP = PB and CS ∥ AP. PBCS is a parallelogram, so PS ∥ BC and PS = BC; hence PQ = BC/2. (ii) For the converse extend PQ to S with PS = BC; BCSP is a parallelogram (equal and parallel), so CS = BP = AP and ∠APQ = ∠CSQ; ΔAPQ ≅ ΔCSQ (AAS) ⇒ AQ = QC.
Examples: rhombus: diagonals perpendicular and bisecting angles; converse: a parallelogram with perpendicular diagonals (or a diagonal bisecting an angle) is a rhombus: true. Rectangle: equal diagonals; converse: a parallelogram with equal diagonals is a rectangle: true (a general quadrilateral is not). Square: a rectangle with perpendicular diagonals / a rhombus with equal diagonals: true.
Hinge on diagonal AC; triangles ABC and ADC keep their own angles at B and D. When ADC is lifted out of the plane, ∠BAD < ∠BAC + ∠CAD and ∠BCD < ∠BCA + ∠ACD (triangle inequality for angles at a vertex), so the sum is less than 360°. As the fold closes, the sum varies continuously and can be made 2° with long thin triangles.
Sliding a copy along the diagonal EO to the next copy gives a shifted copy; the four parallelograms around show an exact match. Repeating along each diagonal gives a 3 × 3 grid of copies meeting only at vertices, and the gaps are congruent to SOME.
(i) Yes iff the largest of a, b, c is less than the sum of the other two (a + b > c for a ≤ b ≤ c). (ii) For sides 2, 5, 11 and x: need 4 < x < 18 (x = 100: no, x = 10: yes, x = 1: no). (iii) A 4-gon exists exactly when the largest side is less than the sum of the other three.
A diagonal joins two non-adjacent vertices. n = 3, 4, 5, 6, 7, 8 give 0, 2, 5, 9, 14, 20. Formula n(n − 3) ÷ 2: each vertex has n − 3 diagonals, each counted twice. Going from n to n + 1 sides adds n − 1 diagonals (n − 2 from the new vertex plus one old side becomes a diagonal): n(n − 3)/2 + (n − 1) = (n + 1)(n − 2)/2.
(n − 2) × 180°: 3 → 180°, 4 → 360°, 5 → 540°, 6 → 720°. Proof: diagonals from one vertex cut the n-gon into n − 2 triangles.
(i) “If PQ ∥ BC and PQ = BC/2 then P and Q are midpoints of AB and AC.” Prove by extending PQ to S so that PS = BC... as in Q15. (ii) Keeping “P is the midpoint of AB”, the converse of “if Q is the midpoint then PQ ∥ BC” is “if PQ ∥ BC then Q is the midpoint of AC”: Theorem 7. (iii) “If P is the midpoint of AB and PQ = BC/2 then Q is the midpoint of AC and PQ ∥ BC?” Not necessarily: the circle of radius BC/2 about P meets line AC in up to two points, and only one is the midpoint.
To prove a method tiles, show (1) the copies fit exactly edge to edge (the half-turn about a side’s midpoint maps a side to itself), (2) around every vertex one copy of each angle meets, so the angles total 360° with no gap or overlap, and (3) the procedure extends in every direction to cover the whole plane.