G3 SEC Mathematics K310 | Learner’s Guide Vol 0174. A geometric proof is not a photograph of a diagram. It is a short chain showing why a relationship must hold. Vectors make that chain easier to build: subtract two position vectors, compare directions, locate a midpoint and show that two sides have the same displacement.
This chapter continues Volume 0170, which introduced position, displacement and magnitude. We now move from finding a vector to using it as proof. Every example is original, and every conclusion is checked against what the vector expression actually establishes. The goal is not longer algebra; it is reasoning that another reader can verify.
1. Check the 2027 K310 boundaries
The official SEAB G3 syllabus directory, updated 26 October 2026, lists Mathematics as K310. The 2027 K310 syllabus places vectors in two dimensions at topic G7. It includes directed line segments, translations, position vectors, magnitudes, sums and differences, scalar multiplication and geometric problems involving vectors.
The assessment scheme consists of Paper 1 and Paper 2, each 2 hours 15 minutes, each worth 90 marks and each contributing 50 per cent. Approved calculators may be used in both. The syllabus expects reasoning and essential working; the last Paper 2 question focuses on a real-world mathematical scenario. These details matter because a vector result in a proof often needs a written justification, not just final coordinates.
This guide stays within the G7 toolkit. It does not require a dot product, cross product, three-dimensional vector geometry or a formal matrix method. You may encounter those in other mathematics courses, but they are not necessary for the two-dimensional SEC reasoning developed here. Use the actual question’s given notation and conditions.
2. One sentence defines displacement
If a point A has position vector a and point B has position vector b from the same origin O, then AB = b − a. That relation is the starting point of almost every problem in this chapter. It says that the directed journey from A to B is the position of B minus the position of A.
For example, A = (2, −1) and B = (8, 5) give AB = (8 − 2, 5 − (−1)) = (6, 6). The journey goes six units right and six units up. Reversing the journey gives BA = (−6, −6). That minus sign is not optional; direction is part of a vector.
Do not confuse the displacement with either point’s position. The vector OA is (2, −1), and OB is (8, 5), but AB is their difference. A clear working line writes the relationship before substituting the numbers.
3. Draw the route, then prove it algebraically
A diagram can suggest that three points lie on a straight line. It cannot establish that a coordinate ratio is exact. Draw a sketch to understand which points appear ordered, then calculate relevant displacement vectors. If one is a scalar multiple of another, they share a line direction or point in opposite directions.
Suppose A = (2, −1), B = (8, 5) and C = (11, 8). We found AB = (6, 6). Now BC = (3, 3), so AB = 2BC. The two displacements are parallel and both point along the line in the same direction. Because B links the journeys from A to C, the points are collinear, with B between A and C.
A strong solution names the vectors and the reason. “AB and BC are proportional; hence A, B and C are collinear” is more informative than writing three coordinate pairs followed by an unexplained check mark.
4. Parallel is not the same as identical
Two non-zero vectors are parallel if one is a scalar multiple of the other. For instance u = (4, −2) and v = (−6, 3) satisfy v = −1.5u. Their directions are opposite, but the corresponding directed segments lie along parallel lines.
Two vectors are equal only if both components match. Equal vectors share magnitude and direction even if drawn at different locations. Parallel vectors need not be equal in length. The difference matters when proving properties of a parallelogram: opposite side vectors may be equal under the correct orientation, not merely parallel.
Keep the zero vector separate. It has no uniquely defined direction, so casually calling it parallel to everything can create confusion. In a geometric proof about sides, check that the points defining a side are distinct where necessary.
5. Read negative scalar multiples correctly
Take u = (2, 5) and v = (−6, −15). Then v = −3u. The lengths are in the ratio 3:1, and the directed vectors point in opposite senses. If they represent two segments on a line, their directions may help identify point order.
A positive multiple such as 0.5u has the same direction; a negative multiple reverses it. The absolute value of the multiplier changes magnitude. This single observation supports many proofs about parallel lines, straight paths and ratios.
Do not confuse the sign of a component with the sign of the scalar multiplier. A vector can contain a negative y-component while still be a positive multiple of another vector. Compare the entire pair of components, not one coordinate in isolation.
6. Collinearity begins with points, not free-floating arrows
To show three distinct points A, B and C lie on one line, use two displacement vectors that connect those actual points, such as AB and AC. If AB = kAC for some real scalar k, then B lies on the line through A and C. The proof is rooted at the same point A, so the direction relationship translates into a statement about the points.
If you merely show that two unrelated segments PQ and RS are parallel, that does not mean P, Q, R and S lie on one line. Parallel segments can belong to different lines. This is a frequent overclaim caused by moving too quickly from vector algebra to geometry.
A well-phrased answer says which two vectors are compared, how the scalar relation was found, and what conclusion follows. State collinearity only when the vectors share the appropriate geometric relationship.
7. The value of k can reveal position along a segment
If AB = kAC with 0 < k 1, B lies beyond C along the ray from A through C. If k < 0, B lies on the opposite side of A.
Suppose A = (1, 2), C = (11, 7) and B = (5, 4). Then AC = (10, 5) and AB = (4, 2) = 0.4AC. Thus B lies two-fifths of the way from A to C. The relation tells us more than “they are on one line”; it gives a position fraction.
This fraction is useful when a question asks how a point divides a segment. Read the direction of the vector before interpreting a ratio. A 0.4 multiple from C toward A would describe a different point.
8. Midpoints are average positions
Let M be the midpoint of segment AB, with positions a and b. Then OM = (a + b)/2. The reason is simple: from A, travel half of AB. Hence OM = a + 0.5(b − a) = 0.5(a + b). This proof shows where the formula comes from instead of treating it as a memory trick.
For A = (2, −1), B = (8, 5), the midpoint is ((2 + 8)/2, (−1 + 5)/2) = (5, 2). The displacement AM is (3, 3) and MB is also (3, 3), confirming that M divides AB into equal directed halves.
Check both components separately. A midpoint is not obtained by averaging x-values while copying one endpoint’s y-value. The formula applies to the full position vector.
9. Prove a proposed midpoint instead of simply accepting it
Suppose A = (1, 4), B = (9, 0) and M = (5, 2). Calculate AM = (4, −2) and MB = (4, −2). Since these directed vectors are equal, the two displacements have the same length and direction and form a single straight journey from A to B. M is the midpoint.
An alternative is to calculate (A+B)/2 = (5, 2) directly. Both are sound for this example. The second is compact; the first can be helpful when the question asks specifically for a vector explanation.
Be careful when a diagram labels a midpoint but the coordinates contradict it. The algebra should decide. In an examination, the phrase “M is the midpoint” may be given as a fact, in which case you use it; when asked to prove it, supply the necessary calculation.
10. A midpoint splits a directed journey into two equal parts
Let AB = (10, −4). If M is the midpoint of AB, then AM = (5, −2) and MB = (5, −2). The two halves are vectors with the same displacement, not opposite vectors. By contrast, MA = (−5, 2) points back toward A.
This is a subtle source of mistakes: students imagine a midpoint as two arrows leaving M in opposite directions, then use those outward arrows in an equality that should involve A→M and M→B. Always write the starting and ending letters in the direction actually needed.
One strong check is vector addition: AM + MB must equal AB. With the numbers above, (5, −2) + (5, −2) = (10, −4). If the proposed halves do not add to AB, the direction or coordinate arithmetic is wrong.
11. Parallelograms combine two directed movements
Let O be the origin and let OA = a, OB = b. If O, A, C and B are vertices of a parallelogram in that order, then OC = a + b. Travelling O→A→C adds a and b. Travelling O→B→C adds b and a. The commutativity of vector addition represents two equivalent routes to C.
This does not mean that any three points automatically define the same intended parallelogram; the vertex order matters. The fourth vertex reached by the two directed side movements is fixed once the adjoining vectors are specified. A careful proof names the order instead of relying on a sketch that may be rotated.
If a = (3, 1) and b = (−2, 4), then C = (1, 5). The edge AC = C − A = (−2, 4) = b and BC = C − B = (3, 1) = a. The opposite sides have equal directed displacement under these chosen orientations.
12. The fourth vertex problem has several possible labels
Given three non-collinear points without a diagram or order, there may be different parallelograms using them as three of the vertices. The question’s naming convention controls which fourth vertex is sought. Do not mechanically add all coordinates as if only one arrangement were possible.
Suppose P, Q and R are consecutive vertices of parallelogram PQRS. Then PQ = SR and QR = PS under matching orientations. One convenient formula is S = P + R − Q, because the two diagonals share a midpoint.
For P=(1,1), Q=(5,2), R=(7,6), this gives S=(1+7−5,1+6−2)=(3,5). Check PQ=(4,1) and SR=R−S=(4,1), while QR=(2,4) and PS=S−P=(2,4). The verification depends on the specified order P,Q,R,S.
13. A diagonal midpoint is a powerful parallelogram test
A quadrilateral whose diagonals bisect each other is a parallelogram. In vector terms, the midpoint of AC is (a+c)/2 and that of BD is (b+d)/2. If these are equal, then a+c = b+d, giving a check on the geometry.
Take A=(0,0), B=(4,1), C=(6,5), D=(2,4). The midpoint of AC is (3,2.5); the midpoint of BD is also (3,2.5). The diagonals share a midpoint, so ABCD is a parallelogram.
These coordinates can also be used to check sides. AB=(4,1) and DC=C−D=(4,1), while AD=(2,4) and BC=C−B=(2,4). Two independent routes strengthen confidence that the geometry has been read correctly.
14. How to prove a line segment is parallel to another
Start by calculating the directed vectors using endpoints. Suppose AB=(6,−4) and CD=(−9,6). The relation CD=−1.5AB establishes that the two segments are parallel, assuming non-zero lengths. The opposite direction does not make the claim false.
If the problem asks for a full geometric conclusion, identify what the vectors represent. Showing AB is parallel to CD does not necessarily make a quadrilateral ABCD a parallelogram; one pair of opposite sides being parallel is insufficient in general. You would need another appropriate property.
A good proof stops at exactly the conclusion justified by the algebra. Avoid writing “therefore the shape is a parallelogram” simply because two computed vectors happen to be parallel. It may be a trapezium instead.
15. Vector ratios and dividing a segment
For A at position a and B at b, a point P one quarter of the way from A to B has position p = a + 0.25(b−a) = 0.75a + 0.25b. This is a weighted position, not an arithmetic average unless the weights are equal.
With A=(−2,4) and B=(10,−4), the displacement AB=(12,−8). One quarter is (3,−2), so P=(1,2). Check that AP=(3,−2) and PB=(9,−6)=3AP. The distance ratio AP:PB is 1:3.
Beware of the word ratio. If AP:PB=1:3, P is one quarter of the total distance from A, not one third. The parts add to four. Diagramming the four equal sections before writing a vector relation can prevent this familiar mistake.
16. Find an unknown using collinearity
Suppose A=(2,1), B=(6,5), and C=(t,9) are collinear. AB=(4,4). For AC=(t−2,8) to be parallel to AB, both components must share the same scalar. The y-component gives multiplier 8/4=2, so the x-component must be 2×4=8. Thus t−2=8 and t=10.
Check using the resulting points. A=(2,1), B=(6,5), C=(10,9) lie on the line that rises one unit for each one unit right. AB=(4,4) and BC=(4,4), so B is actually the midpoint of AC. The collinearity proof has revealed a stronger property.
Do not use only the x-components. Two vectors can have proportional horizontal components but different vertical ratios. Collinearity requires one common scalar across both coordinates.
17. Vector sums can expose a hidden midpoint
Suppose position vectors satisfy OA=a, OB=b and OC=c, with a+c=2b. Rearranging gives b=(a+c)/2. That is precisely the midpoint formula. Thus B is the midpoint of AC, assuming A and C are distinct.
This form often appears in more abstract questions where the diagram uses letters a, b and c rather than numerical coordinates. Learn to read algebra as a geometric statement. The equation does not need a decimal example to be useful.
Conversely, if B is the midpoint of AC, you may immediately write a+c=2b. Do not make a circular proof: if the question asks you to prove the midpoint property, derive this equality from given independent information instead of assuming the conclusion.
18. A triangle’s mid-segment from first principles
Let triangle ABC have position vectors a, b and c. Let M be the midpoint of AB and N the midpoint of AC. Then m=(a+b)/2 and n=(a+c)/2. Therefore MN=n−m=(c−b)/2 = 0.5BC.
The conclusion has two parts. MN is parallel to BC because it is a positive scalar multiple of BC, and its length is half of BC. There is no need to eyeball a diagram. The vector identity establishes both relationships at once.
Notice the subtraction cancels a. That is not a coincidence: moving both midpoints relative to the same vertex A leaves a connecting displacement related to the opposite side. The proof is short because the correct representation exposes the structure.
19. Medians meet at a common point
A triangle’s medians join each vertex to the midpoint of the opposite side. Let A, B and C have positions a, b and c. The midpoint of BC is (b+c)/2. A point two-thirds of the way from A toward that midpoint has position a + (2/3)((b+c)/2−a) = (a+b+c)/3.
The same expression results when starting from B toward the midpoint of AC, or from C toward the midpoint of AB. Thus the three median paths meet at one point, the centroid. This is an elegant geometric extension using only two-dimensional vector sums and scalar multiplication. Students should use it when the question and school work call for this reasoning, not as a formula to force into every SEC problem.
The fraction two-thirds is measured from a vertex toward the opposite midpoint. Reversing it can give the wrong location. A labelled sketch of a median divided into three equal conceptual parts helps anchor the algebra.
20. A worked coordinate centroid check
Let A=(0,0), B=(6,0), C=(0,9). The centroid position is G=((0+6+0)/3,(0+0+9)/3) = (2,3). The midpoint of BC is M=(3,4.5). The displacement AM=(3,4.5), so two-thirds of that is (2,3), reaching G.
From B, the midpoint of AC is N=(0,4.5). The displacement BN=(−6,4.5), and B + (2/3)BN=(6,0)+(−4,3)=(2,3). Both medians reach the same point. The two calculations independently verify the algebra.
Do not assume the centroid has the same property as a circumcentre or an in-centre. Those are different geometric points. The argument here rests specifically on medians and midpoint relations.
21. Translate an entire figure without changing displacement
Let triangle ABC be translated by vector t=(4,−3). The new positions are A’=A+t, B’=B+t and C’=C+t. The displacement A’B’=(B+t)−(A+t)=B−A=AB. Translation changes the position of a figure but not the directed displacement between corresponding points.
Therefore parallelism, lengths and midpoint relationships are preserved by a translation. The proof follows from cancelling the same translation vector. This is the vector language of moving a geometric figure without rotating or resizing it.
A common error is to add t to only one endpoint when computing the translated side. That would change the displacement wrongly. Check that every vertex has moved by the same translation.
22. A route can be different while displacement stays equal
A pupil walks from A=(0,0) to B=(4,0) and then to C=(4,3). The total path length is 4+3=7 units, but the displacement AC is (4,3), with magnitude 5 units. The two quantities answer different questions: how far the pupil travelled versus where the journey ended relative to its start.
This connects geometric vector proof to practical route questions. A path may contain turns, while the net displacement depends only on endpoints. Two different routes from A to C can have equal net displacement but different travel distances.
If asked to recommend the shortest safe route through a building, vector magnitude alone is insufficient. Doors, barriers and permitted corridors matter. This limitation becomes important in an integrated safety or access case. Mathematics models the geometry; the real instructions govern what paths may be used.
23. Three frequent proof failures
Failure one: assuming a diagram is to scale. Points that look aligned may not be. Calculate vectors or use the given geometric facts. Failure two: comparing wrong arrows. AB and BA point opposite ways; an accidental sign reversal can alter the interpretation of a midpoint or ratio.
Failure three: concluding too much. Proportional vectors can establish parallelism, but unrelated parallel segments do not prove four endpoints are collinear. One pair of parallel sides does not automatically give a parallelogram. State the exact geometric property supported.
To repair these failures, write the required property in words first, then ask which vector equality would prove it. This backward planning reduces random calculation and makes the final line easier to justify.
24. How to present an efficient examination proof
A good vector proof normally needs four things: a statement of what must be shown, the correct displacement vectors, a clear equality or scalar relation, and a conclusion naming the relevant geometric property. It does not need theatrical phrases such as “obviously” or “as can be seen”.
For example: “AB = (6,6). BC = (3,3). Since AB = 2BC, the two displacements are parallel and along the same line through B. Therefore A, B and C are collinear.” That sequence gives the reader every necessary reason.
When a proof is more abstract, define the position vectors before using them. For a midpoint: “OM=(OA+OB)/2, so AM=OM−OA=(OB−OA)/2=0.5AB.” Consistent notation protects the algebra from a sign error.
25. Thirty original vector reasoning drills with worked explanations
Try each problem before reading its answer. Use a sketch if it clarifies the point order, then write the vector relation. These exercises train geometric proof from the published G7 toolkit; they are not official SEC question extracts.
1. Displacement
Question. A=(2,−1), B=(8,5). Find AB.
Worked answer. (6,6). Reason. Subtract A from B component by component.
2. Reverse
Question. For the same points, find BA.
Worked answer. (−6,−6). Reason. Reversing direction negates the vector.
3. Net journey
Question. AB=(3,−2), BC=(−1,5). Find AC.
Worked answer. (2,3). Reason. Add consecutive displacements.
4. Scalar multiple
Question. Are u=(4,−2) and v=(−6,3) parallel?
Worked answer. Yes. Reason. v=−1.5u.
5. Equal or parallel
Question. Are u=(2,3) and v=(4,6) equal?
Worked answer. No, but parallel. Reason. v=2u; magnitudes differ.
6. Collinearity
Question. A=(2,−1), B=(8,5), C=(11,8). Are they collinear?
Worked answer. Yes. Reason. AB=(6,6) and BC=(3,3), so AB=2BC.
7. Midpoint
Question. Find midpoint of (2,−1) and (8,5).
Worked answer. (5,2). Reason. Average corresponding coordinates.
8. Midpoint check
Question. A=(1,4), M=(5,2), B=(9,0). Is M the midpoint?
Worked answer. Yes. Reason. AM=(4,−2)=MB.
9. Two halves
Question. AB=(10,−4), M midpoint. Find AM.
Worked answer. (5,−2). Reason. Half the directed displacement.
10. Direction
Question. In the preceding case, find MA.
Worked answer. (−5,2). Reason. Reverse AM, not MB.
11. Ratio point
Question. A=(−2,4), B=(10,−4). Find point P with AP:PB=1:3.
Worked answer. P=(1,2). Reason. P=A+0.25(B−A).
12. Position fraction
Question. If AB=0.4AC, where is B when A and C are distinct?
Worked answer. Two-fifths from A toward C. Reason. The scalar 0.4 lies strictly between zero and one.
13. Beyond endpoint
Question. If AB=1.5AC with distinct A,C, where is B?
Worked answer. Beyond C on ray AC. Reason. The positive multiplier exceeds one.
14. Opposite side
Question. If AB=−0.25AC, which side of A is B?
Worked answer. Opposite C. Reason. A negative scalar reverses direction.
15. Unknown coordinate
Question. A=(2,1), B=(6,5), C=(t,9) collinear. Find t.
Worked answer. 10. Reason. AC must equal 2AB.
16. New midpoint relation
Question. Given a+c=2b, what is B?
Worked answer. The midpoint of AC. Reason. b=(a+c)/2, assuming A and C are distinct.
17. Parallelogram fourth point
Question. P=(1,1), Q=(5,2), R=(7,6) consecutive. Find S.
Worked answer. (3,5). Reason. S=P+R−Q.
18. Opposite sides
Question. For the prior parallelogram, find PQ and SR.
Worked answer. Both (4,1). Reason. P→Q and S→R have equal directed movement.
19. Diagonal midpoints
Question. A=(0,0), B=(4,1), C=(6,5), D=(2,4). Compare AC and BD midpoints.
Worked answer. Both (3,2.5). Reason. Equal diagonal midpoints support a parallelogram.
20. Parallelogram origin
Question. OA=(3,1), OB=(−2,4) are adjoining sides. Find opposite corner C.
Worked answer. (1,5). Reason. OC=OA+OB.
21. Translation
Question. If A=(−1,2) translates by (4,−3), find A’.
Worked answer. (3,−1). Reason. Add the translation components.
22. Translation invariance
Question. After translating both A and B by t, does AB change?
Worked answer. No. Reason. (B+t)−(A+t)=B−A.
23. Magnitude
Question. Find magnitude of (3,4).
Worked answer. 5. Reason. √(3²+4²)=5.
24. Path versus displacement
Question. Walk 4 right and 3 up. Find path length and displacement magnitude.
Worked answer. 7 and 5. Reason. Route distance sums legs; magnitude uses Pythagoras.
25. Midsegment
Question. M,N are midpoints of AB,AC. Express MN in terms of BC.
Worked answer. MN=0.5BC. Reason. Subtract midpoint positions.
26. Centroid
Question. Find centroid of (0,0),(6,0),(0,9).
Worked answer. (2,3). Reason. Average three vertex position vectors.
27. Median fraction
Question. How far from a vertex is a triangle centroid along the median?
Worked answer. Two-thirds toward the opposite midpoint. Reason. The vector expression is vertex+(2/3)(midpoint−vertex).
28. Only one parallel pair
Question. Does one pair of opposite parallel sides prove a quadrilateral is a parallelogram?
Worked answer. No. Reason. A trapezium can have only one such pair.
29. Shared origin
Question. If AB=2AC with A,B,C distinct, are A,B,C collinear?
Worked answer. Yes. Reason. AB and AC share A and one is a scalar multiple of the other.
30. Proof statement
Question. u=(6,−4) and v=(−9,6). State the strongest immediate direction fact.
Worked answer. The corresponding non-zero segments are parallel. Reason. v=−1.5u; collinearity of unrelated endpoints is not established.
26. An integrated proof about a proposed walking route
A school sketch uses a coordinate grid. It marks entrance A=(0,0), a crossing point B=(4,3) and destination C=(8,6). A learner argues that B lies on the straight journey from A to C. Calculate AB=(4,3) and AC=(8,6)=2AB. Thus B is the midpoint of AC, and the three points are collinear.
That conclusion establishes the geometric relationship. It does not prove the route is safe, open, accessible or authorised. A map can show a direct segment crossing a wall or restricted area. The distinction is useful in real-world problems: mathematical displacement identifies a path’s endpoint relationship, while the scenario must tell us whether the path is permitted.
If the proposed corridor route instead goes A→D=(0,6)→C, its total distance is 6+8=14 grid units, whereas the straight displacement magnitude from A to C is √(8²+6²)=10 units. The numerical comparison is valid; the recommendation still depends on barriers and constraints.
27. A progression plan from confident arithmetic to proof
First practise displacement arithmetic until AB=b−a is automatic. Next prove midpoint relations from coordinates rather than simply applying a memorised formula. Then solve collinearity questions with an unknown coordinate. After that, work on parallelograms and midpoint constructions with letters rather than numbers.
In the advanced stage, combine more than one vector relation: use a midpoint to identify a mid-segment, or use diagonal midpoint equality to justify a parallelogram. Always write a final reason in words. A correct scalar multiple hidden in a margin is not a complete geometric explanation.
Use a two-column correction log. In one column record the algebraic error, such as subtracting in the wrong order; in the other record the reasoning error, such as treating parallel segments as collinear points. Repair each with a different type of practice.
28. Advice for parents, students and tutors
When a learner struggles with vectors, do not begin by demanding a longer proof. Ask them to draw one arrow from A to B and describe its horizontal and vertical changes. Then show how an equal displacement can be placed elsewhere on a page without changing the vector. Once that is secure, compare scalar multiples and their geometric meaning.
A tutor can detect misunderstandings with one precise question: “If AB=(3,2), what is BA?” A student who gives the same answer has not yet stabilised direction. Another question, “Does parallel mean equal?” reveals whether magnitude and direction have been separated.
Strong habits from the PSLE Learning Guide—coordinate reading, arithmetic checks and explaining a diagram—support this later formal reasoning. The bridge is deliberate rather than abrupt.
29. Continue through the correct owners
The G3 SEC Learner Guide connects the examination route. For subject depth, use the Mathematics Hub. The Secondary English Hub helps learners explain reasoning clearly, while the Science Hub provides contexts where magnitude, direction and measurement matter.
Read Volume 0170 first if the displacement foundation is weak. For adjacent geometrical reasoning, Volume 0158 builds careful circle measurements, and Volume 0172 distinguishes vector displacement from permitted real-world routes.
For the governing 2027 syllabus, refer to SEAB Mathematics K310. The exercises here are original and belong to a learner’s training, not to a purported official examination paper.
30. The final proof should be smaller than the puzzle
When a problem seems complex, identify the one relation that would settle it. A midpoint invites an average position. Parallel sides invite a scalar multiple. A parallelogram invites equal corresponding displacements or diagonal midpoints. Collinearity invites a pair of connecting vectors with a shared scalar.
Then make every line earn its place. Define the vectors, compute or simplify, and name the geometric conclusion. The examiner does not need a decorative paragraph; the reader needs a defensible chain. The strongest proof is often the one in which no essential step is missing and nothing unsupported has been added.
