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TOPIC 01
Vector proofs in triangles
Turn geometric hypotheses into vector equations; prove midpoint, centroid and orthocenter properties with their assumptions visible.
What you will be able to explain
Choose a convenient origin without changing a geometric claim.
Derive a median identity from a squared norm.
Separate a theorem, its converse and nondegeneracy conditions.
Try it. Leave your reasoning visible.
Use one hint at a time. A correction explains what changed, not just the final answer.
Examples 1–3: study a full demonstration; finish a solution from its first step; then choose a method independently before opening the solution. The remaining exercises may continue after class.
Prove that G=(A+B+C)/3 lies on each median of a nondegenerate triangle.
M=(B+C)/2
Skills and prerequisite lessons
Scope: JM02
Recent attempts (up to five)
Working and explanation
BUILD THE REASONING
Hint 1+
Express G−A using M−A.
Hint 2+
Repeat with the vertices cyclically.
Worked solution+
Rewrite the displacement from A.
G−A=32(M−A)
The same equation holds at B and C.
G−B=32((A+C)/2−B)
Thus all three medians contain G.
The medians concur at G and are divided 2:1 from their vertices.
Checks and common pitfalls: Showing one median alone does not prove concurrency.
Reasoning checklist · self / teacher assessment
The medians concur at G and are divided 2:1 from their vertices.
Showing one median alone does not prove concurrency.
Think first. Reveal a hint when the class is ready.
Teacher preparation and assessment
40-minute lesson
Minutes
Focus
Teaching and student action
5
Prediction and diagnosis
Compare one numerical triangle with a universal claim.
8
Concept derivation
Derive midpoint and median identities with A as origin.
10
Guided application
Use dot products to verify right angles and an orthocenter.
12
Transfer and explanation
Prove the rhombus converse and discuss the missing assumption.
5
Exit evidence
Repair a one-line proof with an unjustified conclusion.
Question sequence
Which origin makes two vertex positions simplest?
Where exactly did the parallelogram assumption enter?
Board development
Arrange givens, vector translation, identity and conclusion in four lines.
For the orthocenter, write one dot-product equation for each altitude and verify the third.
Expected correct and incorrect responses
Students say the diagonals are perpendicular because their lengths match. Expand their dot product to reveal the actual side-length condition.
Correct: “The parallelogram assumption makes its diagonal vectors u+v and v−u.”
Incorrect: “Every quadrilateral with perpendicular diagonals is a rhombus.”
Follow-up questions
Can the centroid formula still be evaluated for collinear points, and does it create a nondegenerate triangle?
Remediation
Start with the midpoint subtraction E−D before introducing a length formula.
Exit check
Write the missing hypothesis in “perpendicular diagonals imply a rhombus”.
Assessment checklist and sample responses
Look for a justified origin, correct directed vectors and a return to the geometric claim. Sample: (u+v)·(u−v)=|u|²−|v|², so perpendicular diagonals force equal adjacent sides in a parallelogram.
Correct: “The parallelogram assumption makes its diagonal vectors u+v and v−u.”
