Build simple geometric proofs with vector identities
Write simple geometric proofs as complete vector identities, with midpoint positions, correctly directed opposite sides and nondegeneracy conditions checked explicitly.
Exam scope and assessment notes
All nine lessons are additional Higher only: section5.1 has seven H5.1A–G rows and no Foundation vector rows. Retain directed endpoint order, signed and zero-vector exceptions, exact components and final-only rounding. Formal lists of two to four coefficients are planar position-vector identities, not independent spatial axes. Complete geometric arguments need their hypotheses and human review; numerical agreement or a text-entry attempt does not approve proof, marks or mastery. Linear4MA1 Issue2 is checked against November2026/June2027 entries; June regional options depend on centre eligibility and modular awards remain separate. Literal Book2 Shape and Space10 H5.1A–G and Hubs167/168/169 links are navigation only. Map427–450 conflicts with proof426/summary449/next450. The Vectors chapter and Hub bodies are unavailable; no Book1 tag, full-book, specialist or formal-assessment approval is inferred.
Before you begin
- Use midpoint positions and directed vector subtraction.
- Recognise nondegenerate triangles, diagonals and boundary order.
- Apply distributive algebra to symbolic planar vectors.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For boundary order A,B,C,D, which directed equality correctly matches one pair of opposite sides in a parallelogram?
What you will learn
- Derive midpoint side identities and connect them to parallel direction and half length.
- Establish a parallelogram from equal opposite directed sides or coincident diagonal midpoints.
- Keep symbolic planar position identities separate from independent-coordinate assumptions.
OAB is a nondegenerate triangle with OA=a, OB=b. M and N are midpoints of OA and OB. Use vectors to prove MN is parallel to AB and half its length.
Draw every requested ordered arrow and enter your own quantities, methods and reasons. Only the stated prompt supplies numerical facts. A symbolic figure is a planar illustration, not extra coordinate givens.
The planar x and y directions use equal drawing scale. Blue arrows are supplied operands; amber arrows are actual learner records; dashed amber is unfinished draft work. A point with coincident tail and head represents the zero vector and has no direction.
This is an illustrative planar pose. Its numerical coordinates, lengths and apparent angles are not additional givens. Formal arrays of two to four symbol coefficients are algebraic identities, never three- or four-dimensional geometry. Only the question states the actual hypotheses.
Actual ordered arrows, zero points and retained drafts do not certify written proofs, marks, mastery or formal assessment.
Construct my actual ordered MN arrow; retain both endpoints, including a coincident zero result.
No actual arrow recorded for this action.
A named pick uses the exact current supplied point or your current clean recorded endpoint. A free click records its labelled micro-unit grid point; exact keyboard fractions remain exact. No endpoint is snapped to a requested result or intersection. A zero vector has coincident endpoints and no defined nonzero direction.
My tail: not placed.
My head: not placed.
Place both your actual ordered tail and head. No endpoint is supplied from a reference. All actual points and raw strings remain yours; no endpoint is invented.
Choose my own illustrative planar source pose
This changes only an illustrative coordinate witness. Enter every permitted origin and source-vector component explicitly before Apply. These coordinates do not become new problem givens or alter formal coefficients. Requested resultant or intersection points are never pose inputs.
My source origin O
My planar source vector a
My planar source vector b
My quantities, directed method and reasons
Enter exactly 2 coefficients in the labelled order a, b, separated by semicolons. These are planar position-vector identities; the list is not a spatial coordinate system.
Enter exactly 2 coefficients in the labelled order a, b, separated by semicolons. These are planar position-vector identities; the list is not a spatial coordinate system.
Use a supported exact rational value or finite sum of single-radical terms where exactness is requested.
Write your actual directed construction, substitutions, hypotheses and reason. A text attempt is not an automatically verified proof.
Write your actual directed construction, substitutions, hypotheses and reason. A text attempt is not an automatically verified proof.
This fixed mathematical tool retains actual planar geometry and requires your complete current attempt. It does not award marks, mastery, automatic proof, physical drawing competence, specialist signoff or formal assessment readiness. Formal coefficients name planar position-vector identities; they do not assert a three- or four-dimensional independent basis.
Turn midpoint information into a directed identity
A midpoint statement gives an exact position-vector equation. Subtract two such midpoint positions to obtain the side joining them, then factor the result as a multiple of the required original side. Name both directed endpoints so that the sign and the geometric correspondence are explicit.
For a nondegenerate triangle, its nonzero sides allow a factor of one half to establish both parallel direction and half length. A complete proof states the identity first and explains those consequences from the factor; a diagram that looks parallel or an isolated numerical example does not prove the general claim.
Match opposite sides with the same traversal sense
For boundary order A,B,C,D, the useful opposite-side equalities are AB=DC and BC=AD. Writing AB=CD reverses one of those directions. Derive the equalities from the supplied position of D rather than assuming the requested parallelogram at the start.
The midpoint of AC has position (a+c)/2 and that of BD has position (b+d)/2. Equality means the diagonals bisect each other in the nondegenerate quadrilateral. This can provide an independent check of the same construction; retain the original noncollinearity conditions so that a collapsed arrangement is not described as a genuine parallelogram.
Use symbolic algebra without inventing independent axes
Symbols a,b,c,d in these proofs denote planar position vectors. An expression involving three or four symbols is still a planar identity, not a three- or four-dimensional coordinate vector. Distributing and cancelling identical vector terms is valid without assuming the position vectors are linearly independent; reverse coefficient comparison would need separately justified independent directions.
For a general quadrilateral, its side midpoints give joining vectors equal to half its diagonals. Prove both pairs of opposite midpoint sides equal, then use nonparallel original diagonals to exclude a collapsed midpoint frame. This establishes a parallelogram, without adding rectangle or rhombus properties. The learner must write and justify the full chain; automatic text acceptance cannot establish that the proof is sound.
Sources for this lesson
- Pearson Mathematics A linear specification · Issue2 · Vectors: Higher-only Foundation statement printed23/physical29; Higher rows printed38/physical44
- Pearson 2026/2027 International GCSE entry manual · regional centres physical3 and Mathematics A physical25
- Pearson literal second-edition book/Hub map · Book2 Shape and Space10 physical15 · navigation only
Pause and explain
P and Q are midpoints of AB and BC. Which identity proves their joining-side relation?
Worked example
OAB is a nondegenerate triangle with OA=a, OB=b. M and N are midpoints of OA and OB. Use vectors to prove MN is parallel to AB and half its length.
Show the worked solution
- Midpoints give OM=a/2 and ON=b/2.
- MN=ON−OM=(b−a)/2=(1/2)AB.
- Nondegeneracy makes AB and MN nonzero. The identity therefore proves MN∥AB in the same direction and |MN|=(1/2)|AB|. The stated midpoint conditions and factor supply the full argument, rather than a measured sketch.
Answer MN=(b−a)/2=(1/2)AB, so MN is parallel to AB and half its length, using the given nondegenerate triangle.
Try a topic test
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A,B,C are noncollinear with position vectors a,b,c. D has position a+c-b, and the boundary order is A,B,C,D. Prove ABCD is a parallelogram by equal opposite directed sides. Also show its diagonal midpoints coincide.
Give me a hint
Expand d in each directed difference; compare AB with DC, not CD, and justify nondegeneracy.
Compare my reasoning
- AB=b−a. Using d=a+c−b gives DC=c−d=c−(a+c−b)=b−a=AB.
- BC=c−b and AD=d−a=(a+c−b)−a=c−b=BC. Both pairs have the required opposite-side traversal sense.
- Since A,B,C are noncollinear, AB and BC are nonzero and nonparallel, so the constructed boundary A,B,C,D is a nondegenerate parallelogram.
- The AC midpoint has position (a+c)/2. The BD midpoint is (b+d)/2=(b+a+c−b)/2=(a+c)/2, so the diagonals bisect each other as an independent check. The symbols are planar position vectors, not three independent spatial coordinates.
AB=DC=b−a and BC=AD=c−b, proving a nondegenerate parallelogram. Both diagonal midpoints equal (a+c)/2.
Look for these in your work
- Derive midpoint side identities and connect them to parallel direction and half length.
- Establish a parallelogram from equal opposite directed sides or coincident diagonal midpoints.
- Keep symbolic planar position identities separate from independent-coordinate assumptions.
ABC is a nondegenerate triangle with positions a,b,c. P,Q,R are midpoints of AB,BC,CA. Prove each side of PQR is parallel to its corresponding side of ABC and half as long. Verify the directed triangle closes.
Give me a hint
Derive all three midpoint differences and match their endpoint orders before summing the closed route.
Compare my reasoning
- The midpoint positions are p=(a+b)/2, q=(b+c)/2 and r=(c+a)/2.
- PQ=q−p=(c−a)/2=(1/2)AC; QR=r−q=(a−b)/2=(1/2)BA.
- RP=p−r=(b−c)/2=(1/2)CB. Each corresponding original side is nonzero because ABC is nondegenerate, giving parallel directions and half lengths.
- PQ+QR+RP=[(c−a)+(a−b)+(b−c)]/2=0. The directed route P→Q→R→P closes. These are identities between planar positions; no independence of a,b,c is assumed.
PQ=(1/2)AC, QR=(1/2)BA and RP=(1/2)CB, proving all three parallel/half-length claims. Their directed sum is zero.
Look for these in your work
- Derive midpoint side identities and connect them to parallel direction and half length.
- Establish a parallelogram from equal opposite directed sides or coincident diagonal midpoints.
- Keep symbolic planar position identities separate from independent-coordinate assumptions.
A simple quadrilateral frame ABCD has four distinct vertices and nonparallel diagonals AC,BD. P,Q,R,S are midpoints of AB,BC,CD,DA. Use position vectors a,b,c,d to prove PQRS is a nondegenerate parallelogram, without assuming ABCD itself is a parallelogram.
Give me a hint
Express each midpoint first; use the original diagonal directions only after both opposite-side identities are derived.
Compare my reasoning
- Write p=(a+b)/2, q=(b+c)/2, r=(c+d)/2, s=(d+a)/2 for the four actual side midpoints.
- PQ=q−p=(c−a)/2 and SR=r−s=(c−a)/2, so PQ=SR.
- QR=r−q=(d−b)/2 and PS=s−p=(d−b)/2, so QR=PS. These are the correctly directed pairs of opposite sides.
- The original diagonals AC and BD are nonparallel and nonzero, so their half-vector directions are also nonparallel. Thus the midpoint frame cannot collapse and PQRS is a nondegenerate parallelogram. No assumption that ABCD is a parallelogram, rectangle or rhombus was used; the four symbols remain planar position vectors.
PQ=SR=(c−a)/2 and QR=PS=(d−b)/2. The nonparallel original diagonals ensure nondegeneracy, so PQRS is a parallelogram without extra properties of ABCD.
Look for these in your work
- Derive midpoint side identities and connect them to parallel direction and half length.
- Establish a parallelogram from equal opposite directed sides or coincident diagonal midpoints.
- Keep symbolic planar position identities separate from independent-coordinate assumptions.
Common mistakes
- Reversing DC to CD in an opposite-side equality.
- Assuming three or four planar position vectors are independent coordinate axes.
- Giving a numerical witness or missing a nondegeneracy hypothesis instead of a general vector proof.
What can you explain now?
If a,b,c are arbitrary planar position vectors, may their coefficients always be compared as three independent coordinate axes?
Compare with the explanation
No. Three arbitrary planar vectors are not an independent three-dimensional basis. Valid symbolic identities can still be derived by expansion and cancellation.
The midpoint proofs use identities valid for the original positions. Reverse coefficient comparison needs an independently justified set of directions, not a count of symbols.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Build a midpoint or parallelogram proof from directed identities, state the nondegeneracy conditions and explain why multiple position symbols do not introduce new spatial dimensions.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.