"Angle Bisector" Proof PPT Courseware

Description

"Angle Bisector" Proof PPT Courseware

Do you remember the properties of points on the angle bisector?

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

Can you prove this conclusion?

Known: As shown in the figure, OC is the bisector of ∠AOB, and P is any point on OC.

PD⊥OA, PE⊥OB, and the vertical feet are D and E respectively.

Prove: PD=PE.

Analysis: To prove PD=PE,

Just prove that they are in △OPD≌△OPB,

And the condition of △OPD≌△OPB

From what is known, it is easy to see that it satisfies the axiom (AAS).

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).

Can you write the converse of "Theorem: The distance from a point on the bisector of an angle to both sides of the angle is equal"?

converse proposition

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

Activities and exploration: Known: As shown in the figure, ∠1=∠2, P is a point on BN, and PD⊥BC is at D, AB+BC=2BD

Verify: ∠BAP+∠BCP=180°

Method summary: (1) When there is an angle bisector (or it is proved to be an angle bisector), draw perpendicular segments to both sides through the points on the angle bisector. When using the judgment or properties of the angle bisector to prove the problem, the problem can often be solved quickly get solution;

(2) When there is a sum-difference relationship between line segments, the truncation method is often used as an auxiliary line to convert the sum-difference relationship into an equal relationship.

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Update Time: 2024-09-15

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

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