"Properties of Angle Bisectors" PPT courseware

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"Properties of Angle Bisectors" PPT courseware

Situation introduction

Tianquan agricultural and sideline products distribution base M is located between the three villages of Lizhuang A, Wangzhuang B, and Zhaozhuang C. Its location is equidistant from the three highways AB, AC, and BC. Can you draw the position of M inside △ABC in the picture?

Move your hands and draw a picture

Ask the students to take out a piece of paper, draw a random corner ∠BAC on the paper, cut it out and fold it in half so that the two sides of the corner overlap, and then unfold it and lay it flat. What did you find?

(1) Thinking: Is an angle an axially symmetrical figure?

If so, find its axis of symmetry.

(2) Conclusion:

An angle is an axially symmetrical figure, and the straight line where the bisector of the angle is located is its axis of symmetry.

Properties of angle bisectors

A point on the bisector of an angle is equidistant from both sides of the angle.

The property is mainly used to prove that two line segments are equal. The premise used is an angular bisector. The key is whether there is "vertical" in the picture.

It is known that: AD is the angle bisector of ∠BAC, P is any point on AD,

Try to explain: PM=PN

Determination of angle bisector:

The point equidistant from the inside of an angle to both sides of the angle is on the bisector of the angle.

The difference and connection between properties and judgment:

The property shows that as long as the point is on the bisector of the angle, its distance to both sides of the angle must be equal, without exception; the judgment reflects that as long as the point is the same distance to both sides of the angle, it must be on the bisector of the angle, and never Missed one.

The former is used to prove that line segments are equal, and the latter is used to prove that angles are equal (angle bisectors)

1. Fill in the blanks:

(1). ∵∠1= ∠2,DC⊥AC, DE⊥AB

∴DC=DE

(The distance from a point on the bisector of an angle to both sides of the angle is equal)

(2). ∵DC⊥AC ,DE⊥AB ,DC=DE

∴∠1= ∠2

(A point equidistant from both sides of an angle is on the angle bisector.)

Known: ∠BAC (as shown in the figure)

Find: OP the angle bisector of ∠BAC

Method: 1. With A as the center of the circle and the appropriate length as the radius, draw an arc, intersect AB at E, and intersect AC at F.

2. Take E and F as the center of the circle respectively, and draw an arc with a length greater than 1/2EF as the radius. The two arcs intersect at point P inside ∠BAC.

3. Make ray AP, and ray OP is what you want.

Proof: Connect PE and PC. We know from the method:

In △AEP and △AFP

AE=AF

PE=PF

AP=AP

∵△AEP≌△AFP(SSS)

∴∠EAP=∠FAP

That is: AP is the angle bisector of ∠BAC.

Question 1. In the second step of the above method, is it okay to remove the condition "longer than 1/2EF"?

[Answer] No. Because if the condition "longer than 1/2EF" is removed, the two arcs may not have an intersection, so the bisector of the angle cannot be found.

Question 2. Is the intersection point of the two arcs made in the second step necessarily inside ∠AOB?

[Answer] If we draw two arcs with E and F as the centers and a length greater than 1/2EF as the radius, the intersection of the two arcs may be inside ∠BAC or outside ∠BAC, and what we are looking for is ∠ The intersection point inside BAC, otherwise the ray obtained by connecting the intersection point of the two arcs and the vertex will not be the bisector of ∠BAC.

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For more information about the "Properties of Angle Bisectors" PPT courseware, please click on the "Properties of Angle Bisectors" ppt tab.

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Update Time: 2024-11-16

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

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