LEARN · EXPLAIN · REVISE

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

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.

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 / 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.

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.”