Class 9 · MathsNCERT 2026 Revised Syllabus

Chapter 12: Quadrilaterals (0904CH12)

Ganita Manjari Part II · 46 questions · 6 exercises

Chapter code: 0904ch12 · Book: Ganita Manjari Part II · Class 9 Maths

Find complete NCERT solutions for Class 9 Maths, Ganita Manjari Part II Chapter 12. Every exercise has a clear, step-by-step answer for chapter code 0904CH12.

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Exercise Set 12.1(5 questions)

Let ABCD be a quadrilateral. (i) List sides adjacent to AB and sides opposite AB. (ii) List angles adjacent to ∠A and opposite ∠A. (iii) Define opposite sides and opposite angles without naming vertices.

School Level Solution

(i) Sides adjacent to AB are AD and BC; the side opposite AB is CD.(ii) The angles adjacent to ∠A are ∠B and ∠D; the angle opposite ∠A is ∠C.(iii) Opposite sides are sides that do not share an endpoint. Opposite angles are angles whose vertices are not endpoints of the same side (the pair of non-adjacent vertex angles).

Precisely define the internal angle of a quadrilateral at a given vertex so that the definition also works for a non-convex quadrilateral.

School Level Solution

At vertex A of quadrilateral ABCD, consider rays AB and AD. The internal angle at A is the angle swept from ray AB to ray AD through the region containing the opposite vertex C (equivalently, the side of the boundary designated as the quadrilateral’s interior). For a convex quadrilateral this is the smaller angle, less than 180°; at a reflex vertex of a non-convex quadrilateral it is the reflex angle, greater than 180°. This specification distinguishes the interior angle from the exterior angle.

In quadrilateral ABCD, suppose (i) AB∥DC; (ii) AB=CD instead; (iii) ∠A=∠C instead. Can ABCD be non-convex in each case?

School Level Solution

(i) If opposite sides AB and DC are parallel in a simple quadrilateral, they lie on distinct parallel lines and the other two sides connect those lines without crossing. The four interior angles are all below 180°; hence ABCD is convex, not non-convex.(ii) Yes. Equal lengths of one pair of opposite sides alone do not force convexity. A concave four-sided figure can be constructed with AB and CD adjusted to have the same length while one interior angle is reflex.(iii) Yes. Equality of just one pair of opposite angles also does not force convexity. For instance, a concave quadrilateral can have ∠A=∠C=60° while its other angles are 190° and 50°; the angle sum is 360° and one angle is reflex.

For non-collinear A, B, C, draw lines AB, BC and CA. For each of the seven regions they create, classify quadrilateral ABCD (vertices in order) as self-intersecting, non-convex or convex.

School Level Solution

The three lines divide the plane into seven regions: the interior of △ABC, three regions immediately across its sides, and three outer vertex regions. For each position of D, connect A–B–C–D–A.• If D is inside △ABC, the quadrilateral is non-convex, with a reflex angle at D. • If D lies across AC, on the side opposite B, ABCD is convex. • If D lies in the region across AB, CD crosses AB; if it lies across BC, AD crosses BC. These are the two self-intersecting cases. • In each of the three outer vertex regions, one of A, B or C lies inside the triangle formed by the other three; the quadrilateral is non-convex.Hence the count is 1 convex, 4 non-convex and 2 self-intersecting.

Can a quadrilateral be both self-intersecting and non-planar?

School Level Solution

A self-intersecting quadrilateral has two non-adjacent side segments that cross. The crossing sides lie on two intersecting lines, which determine a plane containing all four endpoints. Therefore all four vertices are coplanar. A non-planar quadrilateral has non-coplanar vertices, so it cannot also be self-intersecting.

Exercise Set 12.2(4 questions)

True or false: (i) A parallelogram with a right angle is a rectangle. (ii) A rhombus with perpendicular diagonals is a square. (iii) If the diagonals of a parallelogram are equal, it is a rectangle.

School Level Solution

(i) True. Adjacent angles in a parallelogram are supplementary. If one angle is 90°, its adjacent angle is also 90°, and then all four angles are right angles: it is a rectangle.(ii) False. The diagonals of every rhombus are perpendicular, but a rhombus need not be a square. A non-square rhombus is a counterexample. Add the condition that a diagonal’s length equals the other (or that one angle is 90°) to force a square.(iii) True. In a parallelogram, equal diagonals imply two adjacent triangles are congruent in a way that makes an angle a right angle; equivalently, a parallelogram with equal diagonals is a rectangle.

The diagonal AC of parallelogram ABCD bisects ∠A. Show that it bisects ∠C and ABCD is a rhombus.

School Level Solution

Given: ABCD is a parallelogram and ∠DAC=∠CAB. To prove: ∠DCA=∠ACB and ABCD is a rhombus.Since AB∥DC, ∠DCA=∠CAB (alternate interior angles). Given ∠CAB=∠DAC, so ∠DCA=∠DAC. Hence △ADC is isosceles and AD=DC. In a parallelogram, opposite sides are equal: AB=DC and AD=BC. Therefore AB=BC=CD=DA, so ABCD is a rhombus. Also, AD∥BC gives ∠DAC=∠ACB; together with ∠DAC=∠CAB and ∠DCA=∠CAB, we obtain ∠DCA=∠ACB. Hence AC bisects ∠C as required.

Answer Yes/No and supply an extra condition if needed: (i) If the diagonals of ABCD bisect its angles, must it be a rhombus? (ii) If its diagonals bisect each other at right angles, must it be a rhombus? (iii) If its diagonals are equal, must it be a rectangle?

School Level Solution

(i) Yes. If both diagonals bisect the opposite angles, the adjacent right triangles formed by the diagonals are congruent; this gives equal adjacent sides, and the quadrilateral is a rhombus.(ii) Yes. If the diagonals bisect each other, ABCD is a parallelogram. If they also meet at right angles, its adjacent sides form equal right triangles, so adjacent sides are equal; hence it is a rhombus.(iii) No. An isosceles trapezium has equal diagonals but need not be a rectangle. Add the condition that ABCD is a parallelogram; a parallelogram with equal diagonals is a rectangle.

ABCD is a parallelogram with AB≠BC. Show that pairwise intersections of its four angle bisectors form a rectangle. Why assume AB≠BC?

School Level Solution

Proof: In a parallelogram, each pair of adjacent interior angles is supplementary. Let ∠A=α and ∠B=180°−α. Their internal bisectors make an angle α/2+(180°−α)/2=90°. The same calculation holds at each intersection of adjacent angle bisectors. Thus the quadrilateral formed by the four intersection points has four right angles and is a rectangle.If AB=BC, the parallelogram is a rhombus. The bisectors of opposite angles then lie along the same diagonals, causing the “four intersection points” to coincide in pairs, so the inner rectangle degenerates. The assumption AB≠BC excludes that case.

Exercise Set 12.3(5 questions)

In △ABC, P, Q, R are midpoints of AB, AC, BC. (i) Show △PQR is congruent to △QPA and to two other triangles. (ii) Can △ABC be reconstructed from △PQR?

School Level Solution

(i) By the Midpoint Theorem, PQ=BC/2, PR=AC/2 and QR=AB/2. Also AP=PB=AB/2, AQ=QC=AC/2 and BR=RC=BC/2. Thus each of △APQ, △PBR, △QCR and △PQR has side lengths AB/2, AC/2, BC/2 in some order. By SSS, all four are congruent.(ii) Yes, up to congruence: construct a triangle whose sides are twice the corresponding side lengths of △PQR; its side lengths are AB, AC and BC. By SSS, this reconstructed △A′B′C′ is congruent to △ABC.

In △ABC, M and N are midpoints of AB and AC. For any D on BC, show that MN bisects AD.

School Level Solution

Given: M and N are the midpoints of AB and AC; D lies on BC. Let AD meet MN at P. To prove: AP=PD.By the Midpoint Theorem in △ABC, MN∥BC. Since D lies on BC, MN∥BD. In △ABD, M is the midpoint of AB and the line MP (part of MN) is parallel to BD. By the converse of the Midpoint Theorem, P is the midpoint of AD. Thus AP=PD.

ABCD is a trapezium with AB∥DC and AB≠DC. G and H are the midpoints of AC and BD. Prove GH∥AB. Why assume AB≠DC?

School Level Solution

Proof: Draw diagonal AC. In △ABC, G and H are not both midpoints of the same triangle; instead use coordinates or a standard trapezium mid-segment argument. Set A=(0,h), B=(b,h), D=(d,0), C=(d+c,0), where AB=b and DC=c. Then G=((d+c)/2,h/2) and H=((b+d)/2,h/2). They have the same y-coordinate, so GH∥AB∥DC.If AB=DC while the bases are parallel, ABCD is a parallelogram. Its diagonals bisect each other, so G=H and GH is not a non-zero segment. The condition prevents this degenerate case.

Suppose the midpoints of sides AB, BC, CD and DA of a quadrilateral ABCD are P, Q, R and S respectively. (i) Show that PR and QS bisect each other. (ii) Show that if AC = BD, then PR and QS are perpendicular. Is the converse true?

School Level Solution

(i) Given: P, Q, R, S are the midpoints of AB, BC, CD and DA.By the Midpoint Theorem in △ABC and △ADC, PQ ∥ AC and PQ=AC/2; similarly SR ∥ AC and SR=AC/2. Hence PQ ∥ SR and PQ=SR. Likewise, QR ∥ BD and QR=BD/2, while PS ∥ BD and PS=BD/2. Therefore PQRS is a parallelogram (both pairs of opposite sides are parallel). The diagonals of a parallelogram bisect each other, so PR and QS bisect each other.(ii) If AC=BD, then PQ=AC/2=BD/2=QR. Thus the parallelogram PQRS has equal adjacent sides and is a rhombus. The diagonals of a rhombus are perpendicular, so PR ⟂ QS.Converse: Suppose PR ⟂ QS. Since PR and QS bisect each other, PQRS is a parallelogram whose diagonals are perpendicular; hence it is a rhombus. Thus PQ=QR. Since PQ=AC/2 and QR=BD/2, we obtain AC/2=BD/2, so AC=BD. Hence the converse is true.

Suppose PQRS is the Varignon parallelogram of quadrilateral ABCD. (i) Copy only PQRS onto another paper. Show how to reconstruct a congruent copy A′B′C′D′ of ABCD. (ii) Justify that the quadrilateral constructed is congruent to ABCD. (iii) Show that if PQRS is a square, then AC and BD are perpendicular and equal. Prove the converse.

School Level Solution

(i)–(ii) Important mathematical qualification. Copying only the Varignon parallelogram PQRS does not, in general, determine a unique quadrilateral ABCD up to congruence. The construction “copy △APS” would require knowing AP and AS or the position of A, but neither is specified by PQRS alone.A counterexample shows the underdetermination. The square with vertices A=(−2,0), B=(0,2), C=(2,0), D=(0,−2) has midpoint quadrilateral with vertices (−1,1), (1,1), (1,−1), (−1,−1), a square of side 2. The non-square convex quadrilateral A=(−2,0), B=(0,1), C=(2,0), D=(0,−3) has midpoint quadrilateral (−1,0.5), (1,0.5), (1,−1.5), (−1,−1.5), also a square of side 2, yet its side lengths are √5, √5, √13, √13 instead of 2√2, 2√2, 2√2, 2√2. Therefore these two original quadrilaterals are not congruent, although their Varignon parallelograms are congruent. More information (for example, a diagonal-intersection position or one additional appropriate measurement) is needed for a unique reconstruction.(iii) Square condition. By the Midpoint Theorem, adjacent sides of PQRS are parallel to the diagonals AC and BD of ABCD and have lengths AC/2 and BD/2. If PQRS is a square, its adjacent sides are equal and perpendicular, so AC=BD and AC ⟂ BD. Conversely, if AC=BD and AC ⟂ BD, adjacent sides of PQRS are equal and perpendicular. Since PQRS is a parallelogram, it is then a square.

Exercise Set 12.4(2 questions)

Justify why the plane cannot be tiled using only regular pentagons.

School Level Solution

The interior angle of a regular n-gon is (n−2)×180°/n. For n=5, it is 3×180°/5=108°. At a point in a tiling, the angles of the tiles around that point must add to 360°. But 360°/108°=10/3 is not an integer. Three pentagons leave 36° uncovered, while four overlap. Therefore regular pentagons alone cannot tile the plane.

Draw a non-convex 4-gon DART and show how copies of DART tile the plane by the two methods discussed. Which method do you prefer?

School Level Solution

Take a simple concave quadrilateral DART with one reflex angle. Method 1: place one copy beside another by rotating a copy through 180° about the midpoint of a shared edge. The two copies match along that edge; repeat at every exposed edge to extend the pattern without gaps. The four angles of the original quadrilateral sum to 360°, which allows the vertices to close around a point.Method 2: join the midpoints of each copy’s sides. These form a Varignon parallelogram. Tile the plane with copies of that parallelogram, then place the matching DART in each parallelogram cell using the same midpoint arrangement. Both methods give a tiling. The 180°-rotation method is usually easier to construct by hand because each new copy is placed directly against a shared edge.

End-of-Chapter Exercises(24 questions)

Using a fact about parallelograms, show how to tile the plane using any given triangle. (Hint: Use the parallelogram tiling described in the introduction.)

School Level Solution

Take △ABC and a congruent copy. Rotate the copy through 180° about the midpoint M of side BC. The vertices B and C swap, and M is also the midpoint of the segment joining A to its image A′. Thus the diagonals AA′ and BC of quadrilateral ABA′C bisect each other at M, so ABA′C is a parallelogram. Parallelograms tile the plane by translations. Each parallelogram is split by a diagonal into two copies of △ABC; therefore copies of the given triangle tile the plane without gaps or overlaps.

Mark the midpoint of the line drawn on the paper in Fig. 12.33, given that the horizontal lines are equally spaced. Justify your answer.

School Level Solution

The endpoints of the slanted segment lie on the second and sixth horizontal rules. Mark M where the segment crosses the fourth rule, halfway between those two rules. To justify this, draw a perpendicular through M to the horizontal rules, meeting the second and sixth rules at U and V. Equal spacing gives MU=MV. The triangles at the two ends, △MUX and △MVY (where X,Y are the segment endpoints), have a right angle, equal vertical angles at M, and MU=MV; hence they are congruent by ASA. Therefore MX=MY and M is the midpoint of the slanted segment.

A self-intersecting quadrilateral ABCD has AB and CD intersecting at E. Show that ∠A+∠B+∠C+∠D < 360°. Can the sum equal 2°?

School Level Solution

Since E lies on AB and CD, the crossing divides the figure into triangles △EAD and △EBC. Let x=∠AED. The vertically opposite angle ∠BEC also equals x. In △EAD, ∠A+∠D=180°−x; in △EBC, ∠B+∠C=180°−x. Adding gives ∠A+∠B+∠C+∠D=360°−2x<360°, because x>0°. To make the sum 2°, set 360°−2x=2°, giving x=179°. Choose the two crossing lines so that the relevant angle is 179° (the adjacent angle is 1°), and place A,B,C,D on the two lines in the order shown. Then the vertex-angle sum is exactly 2°.

In parallelogram ABCD, points P and Q lie on diagonal BD such that DP=BQ. Show that APCQ is a parallelogram.

School Level Solution

Let O be the intersection of diagonals AC and BD. Since ABCD is a parallelogram, OA=OC and OB=OD. Given DP=BQ, OP=OD−DP and OQ=OB−BQ. Since OD=OB and DP=BQ, OP=OQ. Thus O is the midpoint of both AC and PQ. These are the diagonals of quadrilateral APCQ; a quadrilateral whose diagonals bisect each other is a parallelogram. Hence APCQ is a parallelogram.

A right-triangle-shaped paper cutout is folded so that point A touches point B (Fig. 12.35). Show that the crease can be used to find the midpoint of both AB and AC.

School Level Solution

Let the crease meet AB at M and AC at N. Folding A onto B makes every point of the crease equidistant from A and B; therefore the crease is the perpendicular bisector of AB. Hence AM=MB and crease MN⊥AB. Since △ABC is right-angled at B, BC⊥AB as well, so MN∥BC. In △ABC, a line through midpoint M of AB parallel to BC meets AC at its midpoint (Converse of the Midpoint Theorem). Therefore AN=NC. The same crease identifies the midpoints of both AB and AC.

The medians of △ABC meet at centroid M and divide the triangle into six pieces. (i) Show how to assemble these six pieces into three congruent triangles, and find their side lengths. (ii) Repeat the procedure with each assembled triangle and show that the resulting nine triangles have sides AB/3, BC/3 and AC/3. (iii) Show how to cut and reassemble △ABC into two congruent triangles. (Use median AP and then a median of △APB.)

School Level Solution

(i) Let P,Q,R be the midpoints of BC, CA, AB. The centroid theorem gives AM=2MP, BM=2MQ and CM=2MR. Rotate △MPB by 180° about P. Since PB=PC, B lands on C and M lands on M′ with P the midpoint of MM′. Because B,P,C are collinear, M,P,M′ are collinear; the moved △MPB joins △MPC to form triangle △MM′C. Its sides are MM′=2MP=AM, M′C=MB and MC. Since MB=2MQ and MC=2MR, the lengths are the three centroid-to-vertex segments AM, BM, CM (equivalently, two-thirds of the corresponding medians). Pair the other two pairs around Q and R in the same way. All three resulting triangles have the same three side lengths, so they are congruent by SSS.(ii) In an assembled triangle from part (i), the three medians have lengths BC/2, CA/2 and AB/2: this follows from the parallelograms formed by the half-turn construction and the midpoint relations. Reapplying part (i) makes the sides of each of its three pieces two-thirds of those medians, namely (2/3)(AB/2)=AB/3, (2/3)(BC/2)=BC/3 and (2/3)(AC/2)=AC/3. Doing this for each of the three assembled triangles produces nine congruent triangles with those side lengths.(iii) First cut along median AP, producing △APB and △APC, with P the midpoint of BC. In △APB, let K be the midpoint of AB and draw median PK. Cut along PK. Rotate △PKB by 180° about K: B maps to A, and P maps to P′, with K the midpoint of PP′. The pieces △PKA and the rotated △PKB form △APP′ because P,K,P′ are collinear. Its sides are AP, AP′=BP=PC and PP′=2PK=AC (by the Midpoint Theorem in △ABC). Thus △APP′ and △APC have the same three side lengths AP, PC and AC, so they are congruent by SSS. This gives two congruent triangles after cutting and rearranging the three pieces.

Let ABCD be a trapezium with AB∥DC. E is the midpoint of AD, and a line through E meets BC at F. (i) If EF∥AB, prove that F is the midpoint of BC and that EF=(AB+CD)/2. (ii) If F is the midpoint of BC, prove that EF∥AB.

School Level Solution

(i) Draw diagonal AC and let EF meet it at G. In △ADC, E is the midpoint of AD and EG∥DC, so by the Converse of the Midpoint Theorem G is the midpoint of AC and EG=DC/2. In △CAB, G is the midpoint of CA and GF∥AB, so F is the midpoint of CB and GF=AB/2. Hence EF=EG+GF=DC/2+AB/2=(AB+CD)/2.(ii) Let M be the midpoint of BD. In △DAB, E and M are the midpoints of DA and DB, so EM∥AB. In △DBC, M and F are the midpoints of DB and BC, so MF∥DC∥AB. Since EM and MF pass through M and are both parallel to AB, they are the same line. Therefore E,M,F are collinear and EF∥AB.

The diagonals AC and BD of parallelogram ABCD intersect at O. A line through O meets AB at P and CD at Q. Show that O is the midpoint of PQ.

School Level Solution

Consider △OAP and △OCQ. OA=OC because diagonals of a parallelogram bisect each other. Since AB∥CD, ∠OAP=∠OCQ (alternate interior angles), and ∠AOP=∠COQ (vertically opposite angles). Thus △OAP≅△OCQ by ASA. Hence OP=OQ. As P,O,Q lie on the same line, O is the midpoint of PQ.

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 quadrilaterals AEFD and EBCF.

School Level Solution

By the trapezium midpoint theorem, EF∥AD∥BC and EF=(AD+BC)/2=(3+5)/2=4 cm. Let h be the distance between the two parallel sides. Since E and F are midpoints of the non-parallel sides, EF lies halfway between the bases, so each smaller trapezium has height h/2.Area(AEFD)=½(AD+EF)(h/2)=½(3+4)(h/2)=7h/4. Area(EBCF)=½(EF+BC)(h/2)=½(4+5)(h/2)=9h/4.Therefore the ratio is (7h/4):(9h/4)=7:9.

Consider four points A, B, C, D in a plane with no three collinear. (i) How many different quadrilaterals can be formed if self-intersecting and non-convex quadrilaterals are allowed? (ii) How many are self-intersecting and how many are convex?

School Level Solution

(i) Fix A as the first vertex. The remaining three points can be arranged in 3!=6 orders, but traversing a quadrilateral in reverse gives the same four sides. Thus the number of distinct quadrilaterals is 6/2=3 (for example, ABCD, ABDC and ACBD).(ii) There are two configurations. If all four points are vertices of their convex hull, only the order around the outside is convex; the other two cyclic orders cross, so there is 1 convex and 2 self-intersecting quadrilaterals. If one point lies inside the triangle formed by the other three, each of the 3 orders gives a simple concave quadrilateral; none is convex or self-intersecting.

Suppose P lies on side AB of △ABC and PQ∥BC meets AC at Q. Take AP=1 and AQ=√5. (i) If PB=2, find QC. (ii) If PB=1/3, find QC. (iii) If PB=2/3, find QC. (iv) Show that if AP/PB is rational, then AP/PB=AQ/QC.

School Level Solution

Because PQ∥BC, equally spaced subdivisions on AB produce equally spaced corresponding subdivisions on AC (Intercept Theorem; it can be proved stepwise using the converse of the Midpoint Theorem).(i) AP=1 and PB=2, so AB has 3 equal unit lengths: AP is 1 part and PB is 2 parts. Correspondingly AQ:QC=1:2. Since AQ=√5, QC=2√5.(ii) PB=1/3 and AP=1, so AP:PB=3:1. Thus AQ:QC=3:1, and QC=AQ/3=√5/3.(iii) PB=2/3, so AP:PB=1:(2/3)=3:2. Hence AQ:QC=3:2 and QC=(2/3)AQ=2√5/3.(iv) Let AP/PB=m/n for positive integers m,n. Divide AP into m equal parts and PB into n equal parts, all of common length. Through the division points draw lines parallel to BC. By repeated use of the midpoint/intercept result, the corresponding parts of AC are equal. The m parts comprising AQ therefore have the same unit length as each of the n parts comprising QC, so AQ/QC=m/n=AP/PB.

(i) In parallelogram ABCD, M and N are the midpoints of AB and CD. Show that DM and BN trisect diagonal AC. (ii) Use this result to describe one construction for trisecting any segment PQ. Then give two other constructions: one using the rational-ratio result of Exercise 11 and one using the Centroid Theorem.

School Level Solution

(i) Since M,N are midpoints, MB=AB/2 and DN=CD/2. In a parallelogram AB=CD, so MB=DN; also MB∥DN. Therefore MBND is a parallelogram and DM∥BN. Let DM meet AC at X, and BN meet AC at Y. In △ABY, M is the midpoint of AB and MX∥BY, so X is the midpoint of AY: AX=XY. In △CDX, N is the midpoint of CD and NY∥DX, so Y is the midpoint of CX: CY=YX. Hence AX=XY=YC.(ii), Method 1 (parallelogram): Make PQ a diagonal of a parallelogram PBQD. Let O be the midpoint of PQ; choose B,D on any line through O with OB=OD and construct the parallelogram. Find M and N, the midpoints of PB and QD; the lines DM and BN meet PQ at its trisection points by part (i).Method 2 (parallel intercepts): Draw a ray from P and mark three consecutive equal segments on it, ending at X₃. Join X₃ to Q. Through the first two marks draw lines parallel to X₃Q; by the equal-intercept theorem they cut PQ into three equal parts.Method 3 (centroid): Construct △PBC so that PQ is a median (choose B and C on a line through Q with QB=QC). Draw a second median from B to the midpoint of PC, meeting PQ at centroid G. The centroid theorem gives PG:GQ=2:1, so GQ=PQ/3. Bisect PG to obtain the other trisection point.

Is the Midpoint Theorem for Quadrilaterals (Varignon’s Theorem) true when ABCD is non-convex? What about when it is self-intersecting? (i) Explain. (ii) When does the Varignon parallelogram collapse to a single segment? (iii) Does the reasoning still apply if ABCD is non-planar?

School Level Solution

Let P,Q,R,S be the midpoints of AB, BC, CD, DA. The proof of Varignon’s theorem uses only the Midpoint Theorem in the four triangles formed by the diagonals; it does not use convexity. Thus PQ∥AC∥SR and PQ=SR=AC/2, while QR∥BD∥PS and QR=PS=BD/2. This proves PQRS is a parallelogram for non-convex and self-intersecting figures as well.The parallelogram collapses when its adjacent sides PQ and QR are parallel, that is, when AC∥BD. This occurs in a special crossed configuration.The same vector/midpoint reasoning works in space: opposite midpoint segments remain equal and parallel, so the four midpoint points are coplanar and form a planar parallelogram even if A,B,C,D themselves are non-coplanar.

Let P, Q, R, S lie on sides AB, BC, CD and DA of a quadrilateral ABCD respectively. If PQRS is a parallelogram, must P, Q, R and S be the midpoints of their respective sides? Justify your answer.

School Level Solution

Take square ABCD of side 4. Let P be on AB with AP=1, Q on BC with BQ=1, R on CD with CR=1, and S on DA with DS=1. Each corner triangle, such as △SAP, is right-angled with legs 1 and 3. These four triangles are congruent by SAS, so their opposite hypotenuses satisfy PQ=QR=RS=SP. The four segments also have equal side directions by the symmetry of the square; hence PQRS is a square and therefore a parallelogram. But AP=1 while AB/2=2, so P is not a midpoint; similarly for Q,R,S. Thus the converse is false.

(i) Complete a proof of the Midpoint Theorem by extending PQ beyond Q to S. How far should PQ be extended, and how does this help prove △APQ≅△CSQ? (ii) Give a similar proof of the converse of the Midpoint Theorem (Theorem 7), starting by extending PQ to a suitable point S.

School Level Solution

(i) Given P and Q are the midpoints of AB and AC, extend PQ beyond Q to S so that QS=PQ. Then Q is the midpoint of PS. In △APQ and △CSQ, AQ=CQ, PQ=SQ and ∠AQP=∠CQS (vertically opposite), so the triangles are congruent by SAS. Thus CS=AP=PB and CS∥AB∥PB. Since one pair of opposite sides of PBCS is equal and parallel, PBCS is a parallelogram. Therefore PS∥BC and PS=BC. As PS=PQ+QS=2PQ, we obtain PQ∥BC and PQ=BC/2.(ii) Now P is the midpoint of AB and PQ∥BC. Extend PQ to S so that PS=BC. Since PS∥BC and PS=BC, PBCS is a parallelogram, so CS∥AB and CS=PB=AP. Compare △APQ and △CSQ: AP=CS, ∠PAQ=∠SCQ and ∠APQ=∠CSQ (alternate angles). They are congruent by ASA; hence AQ=CQ, so Q is the midpoint of AC. Also PQ=PS/2=BC/2.

Review the properties of rhombuses, rectangles and squares learned earlier. State some converses and decide whether each converse is true.

School Level Solution

Possible converses (many answers are valid):Rhombus. (a) If the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus — true: bisection makes a parallelogram; perpendicular diagonals make its adjacent sides equal. (b) If both diagonals bisect the angles, it is a rhombus — true, by congruent triangles about each diagonal. (c) If the diagonals are perpendicular, it is a rhombus — false; a kite can have perpendicular diagonals without all four sides equal.Rectangle. (d) If the diagonals are equal, it is a rectangle — false; an isosceles trapezium has equal diagonals but is not a rectangle. (e) If the diagonals are equal and bisect each other, it is a rectangle — true: bisection gives a parallelogram, and a parallelogram with equal diagonals is a rectangle.Square. (f) If diagonals are equal and perpendicular, it is a square — false without bisection (a suitable kite is a counterexample). (g) If diagonals are equal, perpendicular and bisect each other, it is a square — true: the figure is both a rectangle and a rhombus, hence a square.

Show that the sum of the internal angles of a non-planar quadrilateral is always less than 360°. Can a non-planar quadrilateral ABCD have ∠A+∠B+∠C+∠D=2°? (Hint: Treat diagonal AC as a hinge around which △ABC and △ADC rotate.)

School Level Solution

Start with a flat convex quadrilateral ABCD whose angle sum is 360°, and treat AC as a hinge. Keep △ADC fixed while rotating △ABC out of its plane. The angles ∠B and ∠D remain unchanged because each belongs to a rigid triangle. The angles at A and C decrease: as B moves from the opposite side of AC towards D’s side, the distance BD decreases; with AB and AD fixed, the angle between them (∠A) decreases, and similarly ∠C decreases. Therefore, once the quadrilateral is genuinely non-planar, the total is strictly less than 360°.To obtain 2°, choose two very thin triangles on AC, so the limiting sum during near-complete folding can be made smaller than 2°. Initially the flat sum is 360°; during rotation the sum changes continuously. By the intermediate value property, at some non-flat position the sum is exactly 2°.

A third method to tile the plane with a quadrilateral SOME uses translated copies (Fig. 12.42). The diagonals EO and E′O′ are collinear and O=E′. (i) Prove that the slid copy matches SOME exactly using four parallelograms. (ii) Explain how repeating the procedure in both diagonal directions gives a 3×3 arrangement and then tiles the plane.

School Level Solution

(i) Slide a cutout of SOME by the same vector $\overrightarrow{EO}$ , so E maps to E′=O and O maps to O′. Every point shifts by the same vector; hence SS′∥EO, MM′∥EO and each has length EO. Together with E′=O and the collinearity of E,O,O′, the four quadrilaterals ESS′E′, OSS′O′, EMM′E′ and OMM′O′ each have one pair of opposite sides equal and parallel. Each is therefore a parallelogram. Their other sides are equal and parallel too, showing that each side of S′O′M′E′ matches the corresponding side of SOME. Thus the two quadrilaterals coincide exactly after translation.(ii) Translate copies in both directions along diagonals EO and SM. Four new copies fit around the initial quadrilateral, meeting it only at vertices. Repeat outward from each of those copies to make a 3×3 grid and continue. Each blank region between four translated copies is another congruent quadrilateral (possibly rotated 180°), so the copies and the gaps cover the plane exactly without overlap or uncovered region.

Is there a quadrilateral with given side lengths? (i) For positive a≤b≤c, state and justify the condition for a triangle with sides a,b,c. (ii) If three side lengths of a quadrilateral are 2, 5 and 11, can the fourth be 100, 10 or 1? What values are possible? (iii) State a general condition for side lengths a,b,c,d to form a quadrilateral.

School Level Solution

(i) Draw a segment of length c. Draw a circle of radius a at one endpoint and radius b at the other. They meet away from the segment exactly when a+b>c; equality gives a degenerate straight line and a+b<c gives no intersection.(ii) Let the fourth length be x. A quadrilateral can exist only if every side is shorter than the sum of the other three. Thus 11<2+5+x, giving x>4, and x<2+5+11=18. Hence the range is 4<x<18. So 100 is impossible, 10 is possible, and 1 is impossible.(iii) For four positive lengths, a non-degenerate quadrilateral exists exactly when the largest length is less than the sum of the other three. Necessity follows because the other three sides must form a broken path between the endpoints of the longest side. For sufficiency, choose a diagonal length that satisfies the triangle inequalities for two triangles using the four sides, then place those triangles on opposite sides of the diagonal; the resulting quadrilateral has the required four side lengths.

Counting diagonals of a polygon. (i) Define a diagonal and make a table of the number of diagonals for n=3,4,5,6,7,8. (ii) Find a formula for an n-gon, and state how many diagonals are added when the number of sides increases by one.

School Level Solution

A diagonal is a segment joining two non-adjacent vertices. The counts are: n=3→0, n=4→2, n=5→5, n=6→9, n=7→14, n=8→20.From each of n vertices, diagonals can be drawn to n−3 other vertices (exclude itself and its two neighbours). This gives n(n−3) diagonal-ends, but each diagonal has two ends, so $D(n)=\frac{n(n-3)}2$ .Increasing the number of sides from n to n+1 adds n−1 diagonals: the new vertex joins to n−2 old non-neighbours, and the old side between its two neighbours becomes one new diagonal. Algebraically, D(n+1)−D(n)=[(n+1)(n−2)−n(n−3)]/2=n−1.

Sum of angles of a polygon. Find the sums for n=3,4,5,6,7,8 and prove a formula for the interior-angle sum of a planar, non-self-intersecting n-gon.

School Level Solution

Join one vertex of a convex n-gon to all its non-adjacent vertices. The diagonals divide the polygon into n−2 non-overlapping triangles. Each triangle has angle sum 180°, and together their angles make precisely the polygon’s internal angles. Therefore the sum is $(n-2) imes180^\circ$ . For n=3,4,5,6,7,8 the values are 180°, 360°, 540°, 720°, 900°, 1080°.

Multiple converses to the Midpoint Theorem. Let P lie on AB and Q on AC in △ABC. Use (P MID): P is midpoint of AB; (Q MID): Q is midpoint of AC; (PRLL): PQ∥BC; (HALF): PQ=BC/2. (i) Prove that (PRLL)+(HALF) imply (P MID)+(Q MID). (ii) Write the converse of the Midpoint Theorem while retaining P MID as an assumption. (iii) If P MID and HALF hold, can Q MID and/or PRLL be concluded?

School Level Solution

(i) Since PQ∥BC, △APQ∼△ABC. The ratio PQ/BC=1/2 therefore gives AP/AB=AQ/AC=1/2. Thus AP=PB and AQ=QC, so P and Q are the midpoints.(ii) The statement is: “If P is the midpoint of AB and the line through P parallel to BC meets AC at Q, then Q is the midpoint of AC.” By Theorem 7, PQ=BC/2 as well.(iii) Neither conclusion follows. A counterexample: take A=(0,0), B=(2,0), C=(3,1) and P=(1,0), which is the midpoint of AB. Let Q=(0.3,0.1), on AC. Then PQ²=(0.3−1)²+(0.1−0)²=0.50, so PQ=√0.5=√2/2; BC=√2 and hence PQ=BC/2. But Q is not the midpoint of AC (that would be (1.5,0.5)) and the slope of PQ is −1/7 while the slope of BC is 1, so PQ is not parallel to BC. Thus P MID and HALF alone imply neither Q MID nor PRLL.

Validity of tiling methods. Give a reasoned proof that each of the three methods for tiling the plane with a given quadrilateral covers the whole plane without gaps or overlaps.

School Level Solution

Method 1: half-turn about a side midpoint. Rotate a copy of the quadrilateral by 180° about the midpoint of a side. The side endpoints swap, so the two copies share the complete side exactly. Repeating at each exposed side gives an edge-to-edge pattern. Around each meeting vertex the four interior angles of the quadrilateral occur once each; their sum is 360°, so there is neither a gap nor an overlap.Method 2: fit four copies around a vertex. Place four copies so that their four interior angles meet at one point. A quadrilateral’s angles sum to 360°, so they fill the full turn exactly. Continue the pattern by matching copies along full sides using half-turns; every local gap is filled and no two interiors overlap.Method 3: translation grid. The construction in Exercise 18 shows that a copy translated by a diagonal vector matches exactly. Repeating the same translation in both diagonal directions creates a grid of translated copies. Each remaining gap is a congruent, turned copy of the original quadrilateral, so the copies and gaps fill the plane without overlap.

What fraction of the square is shaded in Fig. 12.43?

School Level Solution

Let the square have side 1, with corners (0,0), (1,0), (1,1), (0,1). The marked points are the midpoints of all four sides. The four lines shown have equations (after matching them to the figure): y=1−2x, y=1/2+x/2, y=2−2x and y=x/2. Their consecutive intersections forming the shaded parallelogram are (1/5,3/5), (3/5,4/5), (4/5,2/5) and (2/5,1/5). Adjacent side vectors can be taken as (2/5,1/5) and (1/5,−2/5). The parallelogram area is the absolute determinant: |(2/5)(−2/5)−(1/5)(1/5)|=|−4/25−1/25|=5/25=1/5. The square’s area is 1, so the shaded fraction is 1/5.

In-text and Think and Reflect(6 questions)

Can every quadrilateral tile the plane? Explain why copies of any simple quadrilateral can cover the plane without gaps or overlaps.

School Level Solution

Take a copy of the quadrilateral and rotate a matching copy through 180° about the midpoint of one side; that side matches exactly because its endpoints are exchanged. Repeat this construction on the other sides and continue the pattern. At a vertex, the four interior angles of the quadrilateral (one of each angle) add to 360°, so the copies fit around a point without a gap or overlap. Thus any simple quadrilateral admits a tiling.

Give a precise definition of a (simple planar) quadrilateral that excludes crossed or overlapping figures.

School Level Solution

The definition must exclude non-planar figures, crossed/self-intersecting boundaries, overlapping sides and degenerate figures where three consecutive vertices lie on the same straight line. Requiring a simple closed polygon with four distinct coplanar vertices and four non-zero straight edges rules these out.

In a quadrilateral, if both pairs of opposite angles are equal, what can be said about adjacent angles? Is the converse true?

School Level Solution

Let ∠A = ∠C and ∠B = ∠D. Since the angle sum is 360°, 2∠A + 2∠B = 360°, hence ∠A + ∠B = 180°; adjacent angles are supplementary. Conversely, if each adjacent pair is supplementary, then ∠A + ∠B = 180° and ∠B + ∠C = 180° imply ∠A = ∠C; similarly ∠B = ∠D. A quadrilateral with both pairs of opposite angles equal is a parallelogram.

State and prove the converse of the midpoint theorem: in △ABC, P is the midpoint of AB and PQ ∥ BC meets AC at Q. Show that Q is the midpoint of AC.

School Level Solution

Because PQ ∥ BC, △APQ ∼ △ABC by AA. Therefore AP/AB = AQ/AC. Since P is the midpoint, AP = AB/2, so AQ/AC = 1/2, giving AQ = AC/2. Hence QC = AC − AQ = AC/2 = AQ, and Q is the midpoint of AC.

State and prove the Midpoint Theorem: in △ABC, P and Q are the midpoints of AB and AC. Prove that PQ ∥ BC and PQ = BC/2.

School Level Solution

Extend PQ beyond Q to R so that QR=PQ. Since Q is the midpoint of AC, AQ=CQ; also ∠AQP=∠CQR (vertically opposite angles) and PQ=QR. Thus △AQP≅△CQR by SAS, giving AP=CR and AP∥CR. Since P is the midpoint of AB, AP=PB and A,P,B are collinear. Therefore PB and CR are equal and parallel, so PBCR is a parallelogram. Hence PR∥BC and PR=BC. As P,Q,R are collinear and PQ=QR, PR=2PQ. Thus PQ∥BC and PQ=BC/2.

Give the alternative proof of the Centroid Theorem suggested in the text. Let M be the intersection of medians CP and BQ. Extend AM to S so that M is the midpoint of AS. If R=AM∩BC, prove that R is the midpoint of BC.

School Level Solution

Let P and Q be the midpoints of AB and AC. By construction, M is also the midpoint of AS. In △ABS, P and M are midpoints of AB and AS, so PM∥BS by the Midpoint Theorem. Since C,M,P are collinear, PM lies on CM; hence CM∥BS. In △ACS, Q and M are midpoints of AC and AS, so QM∥CS. Since B,M,Q are collinear, QM lies on BM; hence BM∥CS. Therefore BSCM is a parallelogram. Its diagonals BC and SM bisect each other. Since S,M,A are collinear, the intersection R of AM and BC is the midpoint of BC. Hence AM is the third median.

Other Chapters

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