"Perpendicular bisectors of line segments" PPT courseware 9

Description

"Perpendicular bisectors of line segments" PPT courseware 9

learning target

1. Be able to use the ruler and compass method to draw the perpendicular bisector of a known line segment, and be able to prove its correctness.

2. Experience the process of exploration to prove the theorem of the properties of perpendicular bisectors of line segments and its inverse theorem, and further develop students’ awareness and ability of reasoning and proof.

3. Be able to use the property theorem of the perpendicular bisector of a line segment and its converse theorem to prove relevant conclusions, and understand that the perpendicular bisectors of the three sides of a triangle intersect at one point, and the distance from this point to the three vertices of the triangle is equal.

review review

What is the perpendicular bisector of a line segment? Is the line segment an axially symmetrical figure? If so, what is its axis of symmetry?

Explore new knowledge

1. How to draw the perpendicular bisector of a line segment?

(1) Origami method

(2) Definition method

(3) Rule and compass drawing method

Rule-and-rule practice

practice:

1. With points A and B as the center points and a radius greater than AB, draw arcs to intersect at points E and F.

2. Draw a straight line through points E and F.

Then the straight line EF is the perpendicular bisector of the line segment AB (Figure 16-11)

2. Why is the straight line EF drawn in this way the perpendicular bisector of the line segment AB? Assume that the straight line EF intersects AB at point O. Please give a proof based on the determination theorem of congruence of triangles.

Prove: Connect AE, AF, BE, BF∴AE=BE=AF=BF (equal circles or the radii of the same circles are equal)

In △AEF and △BEF

∵AE=BE (certified)

AF=BF (certified)

EF＝EF (common edge)

∴ △ AEF≌ △ BEF (SSS)

∴ ∠AEO= ∠ BEO (corresponding angles of congruent triangles are equal)

In △ AEO and △ BEO

∵ AE＝BE (certified)

∠ AEO＝ ∠ BEO (certified)

EO=EO (common edge)

∴ △ AEO≌ △ BEO (SAS)

∴ AO=BO (corresponding sides of congruent triangles are equal)

∠ AOE＝ ∠ BOE (corresponding angles of congruent triangles are equal)

∵ ∠ AOE+ ∠ BOE＝180 (definition of adjacent supplementary angles)

∴ ∠ AOE＝ ∠ BOE＝90 (property of equality)

∴EF⊥AB (vertical definition)

∴EF is the perpendicular bisector of line segment AB (definition of perpendicular bisector of line segment)

Property Theorem: A point on the perpendicular bisector of a line segment is equidistant from both ends of the line segment.

Class summary

This lesson focuses on two knowledge points:

1. The points on the perpendicular bisector of the line segment are equidistant from both ends of the line segment.

2. A point equidistant from both ends of a line segment is on the perpendicular bisector of the line segment.

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File Info

Update Time: 2024-10-03

This template belongs to Mathematics courseware Hebei Education Edition Eighth Grade Mathematics Volume 1 industry PPT template

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