"Simple Axisymmetric Figures" Axisymmetric PPT Courseware 3

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"Simple Axisymmetric Figures" Axisymmetric PPT Courseware 3

Situational question one

Without using tools, please divide a corner made of paper into two equal corners. What can you do?

Situational question two

For this kind of foldable angle, we can use the angle bisector of the folding method. How to get the angle bisector of the angle that cannot be folded?

There is a simple instrument that bisects an angle (as shown in the picture), where AB=AD, BC=DC, put point A at the vertex of the angle, AB and AD draw a ray AE along AC, AE is the bisector of ∠BAD, why?

How to find the bisector of an angle using a ruler and compass

practice:

1． Take O as the center of the circle, draw an arc with an appropriate length as the radius, intersect OA at M, and intersect OB at N.

2. Take M and N as the center of the circle respectively. For any length greater than 1/2MN, draw an arc with the radius. The two arcs intersect at C inside ∠AOB.

3. Make ray OC． Then the ray OC is what is required.

Situational question three

Fold ∠AOB in half, then fold it into a right triangle (making the first crease a hypotenuse), then unfold it and observe the three creases formed by the two folds. What conclusion can you draw?

Explore the properties of angle bisectors

You can take a look at it. The first crease is the bisector OC of ∠AOB, the two creases PD formed by the second fold, and PE are the distances from one point on the bisector of the angle to both sides of ∠AOB. These two distances are equal. .

Points on the bisector of an angle are equidistant from both sides of the angle.

Known: As shown in the figure, OC is the bisector of ∠AOB, point P is on OC, PD⊥OA, PE⊥OB, and the vertical feet are D and E respectively.

Verify: PD=PE

Proof: ∵ PD⊥OA, PE⊥OB (known)

∴∠PDO=∠PEO=90 (vertical definition)

In △PDO and △PEO

∠ PDO= ∠ PEO

∠ AOC= ∠ BOC

OP=OP

∴ △ PDO≌ △ PEO (AAS)

∴ PD=PE (corresponding sides of congruent triangles are equal)

practice

1. As shown in the figure, ∵ OC is the bisector of ∠AOB, and PD⊥OA, PE⊥OB

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

2. In Rt△ABC, BD is the angle bisector, DE⊥AB, and the vertical foot is E. Are DE and DC equal? Why?

3. As shown in the figure, OC is the bisector of ∠AOB, point P is on OC, PD⊥OA, PE⊥OB, and the vertical feet are D and E respectively, PD=4cm, then PE=__________cm.

4. Given that in △ABC, ∠C=900, AD bisects ∠CAB, BC=8, BD=5, what is the distance from point D to AB?

Summary and expansion

What knowledge did we learn in this lesson?

1. The ruler and compass drawing method of "drawing the bisector of a known angle";

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

Geometric language:

∵ OC is the bisector of ∠AOB,

Also PD⊥OA,PE⊥OB

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

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Update Time: 2024-09-08

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