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Vector proofs in triangles

Read the idea, work independently, then explain what changed.

高一必修第二册(A版).pdf · 6.4 · PDF 45 / printed page 38

Revisit first: Coordinates, bases and division Dot products and constrained projection

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.

An origin turns a triangle into two vectors

A proof should explain why every triangle with the assumptions works, rather than verify one drawing.

Set A as origin and write B=u,C=v. Midpoints become averages, side lengths become norms and perpendicularity becomes a zero dot product.

M=(u+v)/2,\quad G=(u+v)/3

For midpoints D=u/2 and E=v/2, subtract their positions.
$DE=(v-u)/2=BC/2$
Expand a median norm and eliminate the dot product using the opposite side.
$4AM^2=2|u|^2+2|v|^2-|u-v|^2$

When it applies

A triangle needs noncollinear vertices.
A rhombus converse here assumes a nondegenerate parallelogram.

Counterexample: Perpendicular diagonals in an arbitrary quadrilateral do not force all four sides equal.

Triangle sketch → two position vectors → scalar identity → geometric conclusion.

PREDICT → EXPLORE → EXPLAIN → TRANSFER

Make a prediction, then explore the relationship.

Lesson question: Predict whether the medians still meet as the third vertex crosses the baseline.

Model exploration: predict which displayed result changes with the parameters, check the values, and compare the observation with the lesson question.

The medians contain G=(A+B+C)/3 in ratio 2:1. At C_y=−1, the positions are collinear and area is zero.

Numerical data for the diagram; printable
Quantity	Value
G	0.33333333, 0.33333333
Area / 面積	12
Collinear / 共線	false / 否
AG/GM	2

Explain: The affine centroid identity remains true at collinearity, but triangle area becomes zero.

Transfer: Choose a degenerate setting and explain which algebraic identities survive and which triangle statements lose their usual meaning.

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.

01 / Foundation#Worked example

D,E are midpoints of AB,AC in a nondegenerate triangle. Prove DE=BC/2 as directed vectors.

AB=u,\ AC=v

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

Place A at the origin.

Hint 2

Subtract D from E.

Worked solution

Midpoint positions halve the side vectors.
$AD=u/2,\ AE=v/2$
Subtract and identify BC.
$DE=(v-u)/2=BC/2$

DE is parallel to BC and has half its length.

Checks and common pitfalls: The subtraction order determines the directed sign.

Reasoning checklist · self / teacher assessment

DE is parallel to BC and has half its length.
The subtraction order determines the directed sign.

Think first. Reveal a hint when the class is ready.

02 / Standard#Worked example

Derive the cosine rule for the side opposite angle A.

AB=u,\ AC=v,\ |u|=c,\ |v|=b,\ |BC|=a

First step provided — complete the reasoning:Use the dot-product expansion.

a^2=|v-u|^2=b^2+c^2-2u\cdot v

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

BC=v−u.

Hint 2

Expand its squared norm.

Worked solution

Use the dot-product expansion.
$a^2=|v-u|^2=b^2+c^2-2u\cdot v$
Insert the included-angle relation.
$u\cdot v=bc\cos A$

a²=b²+c²−2bc cos A.

Checks and common pitfalls: The angle must be between the two given side vectors.

Reasoning checklist · self / teacher assessment

a²=b²+c²−2bc cos A.
The angle must be between the two given side vectors.

Think first. Reveal a hint when the class is ready.

03 / Transfer#Worked example

Derive the squared length of the median AM from side lengths.

M=(B+C)/2,\ AB=c,\ AC=b,\ BC=a

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

Use AM=(u+v)/2.

Hint 2

Eliminate u·v using |u−v|².

Worked solution

Set A as origin, u=AB and v=AC; combine the sum and difference norm identities.
$|u+v|^2=2b^2+2c^2-a^2$
Scale the median vector by one half.
$AM^2=(2b^2+2c^2-a^2)/4$

AM²=(2b²+2c²−a²)/4.

Checks and common pitfalls: Halving a vector divides its squared norm by four.

Reasoning checklist · self / teacher assessment

AM²=(2b²+2c²−a²)/4.
Halving a vector divides its squared norm by four.

Think first. Reveal a hint when the class is ready.

04 / Foundation#Your turn

Find the area of triangle ABC.

A=(0,0),B=(6,0),C=(2,4)

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Use AB as horizontal base.

Hint 2

Perpendicular height is the y-coordinate.

Worked solution

Identify base and height.
$b=6,\ h=4$
Take half the rectangle area.
$S=6\cdot4/2=12$

Area 12.

Checks and common pitfalls: AC is not the perpendicular height.

Think first. Reveal a hint when the class is ready.

05 / Foundation#Your turn

Two sides are 3 and 5 with included angle 60°. Find the square of the opposite side.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Use the cosine rule.

Hint 2

cos60°=1/2.

Worked solution

Insert the included angle.
$a^2=3^2+5^2-2(3)(5)\cos60^\circ$
Evaluate.
$a^2=34-15=19$

The squared side length is 19.

Checks and common pitfalls: The question asks for the square, not √19.

Think first. Reveal a hint when the class is ready.

06 / Foundation#Your turn

The median AM has length 9 and G is the centroid. Find AG.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The vertex-to-centroid portion is two-thirds.

Hint 2

Apply the ratio to the whole median.

Worked solution

Use the centroid position along AM.
$AG=\frac23 AM$
Substitute the median length.
$AG=6$

AG=6.

Checks and common pitfalls: The shorter third is GM, not AG.

Think first. Reveal a hint when the class is ready.

07 / Foundation#Your turn

Decide whether the angle at A is right.

A=(1,1),B=(3,2),C=(0,3)

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

Form both vectors from A.

Hint 2

Test their dot product.

Worked solution

Compute the side vectors.
$AB=(2,1),\ AC=(-1,2)$
Both are nonzero and orthogonal.
$AB\cdot AC=-2+2=0$

Yes, ∠A=90°.

Checks and common pitfalls: Using BC instead of AC would test a different angle.

Reasoning checklist · self / teacher assessment

Yes, ∠A=90°.
Using BC instead of AC would test a different angle.

Think first. Reveal a hint when the class is ready.

08 / Standard#Your turn

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

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.
$G-A=\frac23(M-A)$
The same equation holds at B and C.
$G-B=\frac23((A+C)/2-B)$
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.

09 / Standard#Your turn

Find the positive area of the triangle whose side vectors are u,v.

u=(1,3),\ v=(4,2)

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

A determinant gives signed parallelogram area.

Hint 2

Take absolute value and halve.

Worked solution

Calculate the determinant.
$\det(u,v)=2-12=-10$
Convert signed parallelogram area to triangle area.
$S=|-10|/2=5$

Area 5.

Checks and common pitfalls: Negative orientation is not negative physical area.

Think first. Reveal a hint when the class is ready.

10 / Standard#Your turn

Prove the diagonals of a rhombus are perpendicular using its adjacent side vectors.

AB=u,\ AD=v,\ |u|=|v|

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

The diagonals have directions u+v and v−u.

Hint 2

Use equal side lengths after expanding.

Worked solution

The mixed products cancel.
$(u+v)\cdot(v-u)=|v|^2-|u|^2=0$
A nondegenerate rhombus has two nonzero diagonals.

The diagonals are perpendicular.

Checks and common pitfalls: Parallel sides alone give a parallelogram, whose diagonals need not be perpendicular.

Reasoning checklist · self / teacher assessment

The diagonals are perpendicular.
Parallel sides alone give a parallelogram, whose diagonals need not be perpendicular.

Think first. Reveal a hint when the class is ready.

11 / Standard#Your turn

A triangle has AB=5, AC=5 and BC=6. Find the median from A.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The isosceles median bisects the base perpendicularly.

Hint 2

Use a right triangle with hypotenuse 5.

Worked solution

Half the base is 3.
$AM^2=5^2-3^2$
Take the positive length.
$AM=4$

AM=4.

Checks and common pitfalls: The median formula and the isosceles right-triangle method agree.

Think first. Reveal a hint when the class is ready.

12 / Transfer#Your turn

A nondegenerate parallelogram has perpendicular diagonals. Prove it is a rhombus.

AB=u,\ AD=v

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

Start from the conclusion supplied as a hypothesis.

Hint 2

Recover equality of squared side lengths.

Worked solution

Translate perpendicularity into algebra.
$0=(u+v)\cdot(v-u)=|v|^2-|u|^2$
Positive lengths are equal; opposite sides already match.
$|u|=|v|$

All four sides are equal, so it is a rhombus.

Checks and common pitfalls: The converse needs the parallelogram assumption; arbitrary perpendicular-diagonal quadrilaterals are insufficient.

Reasoning checklist · self / teacher assessment

All four sides are equal, so it is a rhombus.
The converse needs the parallelogram assumption; arbitrary perpendicular-diagonal quadrilaterals are insufficient.

Think first. Reveal a hint when the class is ready.

13 / Transfer#Your turn

Find the orthocenter of triangle ABC, and verify all three altitude conditions.

A=(0,0),B=(6,0),C=(2,4)

Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1

The altitude from C is vertical.

Hint 2

Use a dot product for the altitude from A.

Worked solution

AB is horizontal, so CH⊥AB requires H=(2,h); now impose AH⊥BC.
$(2,h)\cdot(-4,4)=0\Rightarrow h=2$
Check the remaining altitude.
$BH=(-4,2),\ AC=(2,4),\ BH\cdot AC=0$

H=(2,2); CH⊥AB, AH⊥BC and BH⊥AC.

Checks and common pitfalls: An altitude is perpendicular to the opposite side, not an adjacent side.

Reasoning checklist · self / teacher assessment

H=(2,2); CH⊥AB, AH⊥BC and BH⊥AC.
An altitude is perpendicular to the opposite side, not an adjacent side.

Think first. Reveal a hint when the class is ready.

End-of-lesson check and correction

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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.”