"Angle Bisector" Proof PPT Courseware 2

Description

"Angle Bisector" Proof PPT Courseware 2

Use ruler and compass to find bisector of angle

Known: ∠AOB, as shown in the figure.

Find the solution: ray OC, so that ∠AOC=∠BOC.

practice:

1. Intercept OD and OE on OA and OB respectively, so that OD=OE.

2. Take points D and E as the center points and draw arcs with a length greater than DE/2 as the radius. The two arcs intersect at point C in ∠AOB.

3. Make ray OC.

Then the ray OC is the bisector of ∠AOB.

theorem

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

As shown in the figure, ∵OC is the bisector of ∠AOB,

P is any point on OC, PD⊥OA, PE⊥OB, and the vertical feet are D and E respectively (known)

∴PD=PE (the distance from the point on the bisector of the angle to both sides of the angle is equal).

Summary and expansion

Theorem: The distance from a point on the bisector of an angle to both sides of the angle is equal.

Converse Theorem: A point inside an angle and equidistant from both sides of the angle lies on the bisector of the angle.

Theorem: The three angle bisectors of a triangle intersect at a point, and the distances from this point to the three sides are equal (this intersection point is called the incenter of the triangle).

The point where an interior angle of a triangle intersects with the bisectors of two non-adjacent exterior angles is called the circumcenter of the triangle. There are three such points.

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Update Time: 2024-09-01

This template belongs to Mathematics courseware Beijing Normal University Ninth Grade First Volume Mathematics industry PPT template

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