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Authoritative PPT Summary
"Angle Bisector" PPT courseware
learn new knowledge
We once used the origami method to obtain the properties of the angle bisector and the points on the angle bisector. Do you still remember the properties of the points on the angle bisector?
Points on the bisector of an angle are equidistant from both sides of the angle.
Combined with the proof method of the theorem we learned earlier, can you write the proof process of this property?
Known: 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.
Prove: PD=PE.
prove:
∵ OC is the bisector of ∠AOB
∴ ∠1= ∠2
∵ PD⊥OA,PE⊥OB
∴ ∠PDO=∠PEO
∵OP=OP
∴ △OPD≌△OPE (AAS).
∴PD=PE
Theorem: The distance from a point on the bisector of an angle to both sides of the angle is equal.
Geometric language, 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).
Tip: This conclusion is one of the bases often used to prove that two line segments are equal.
Can you write the converse of "The above 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.
Use a ruler and compass to find the bisector of the 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 of the circles and draw arcs with a radius greater than the length. The two arcs intersect at point C in ∠AOB.
3. As ray OC, ray OC is the bisector of ∠AOB.
Construct the three angle bisectors of the triangle:
Observe these three angle bisectors, what do you find?
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)
Challenge yourself
1. As shown in the figure, AD and AE are the interior angle bisectors and exterior angle bisectors of ∠A in △ABC respectively. What is their relationship?
2. As shown in the figure, a target is in area A, equidistant from the road and railway, and 500m from the intersection of road and railway. Mark its location on the map (scale 1:20000).
3. As shown in the figure, find a point P such that PC=PD, and the distance from point P to both sides of ∠AOB is equal.
4. Known: As shown in the figure, ∠C=900, ∠B=300, AD is the angle bisector of Rt△ABC.
Verify: BD=2CD.
Review and summary
1. Theorem:
A point on the bisector of an angle is equidistant from both sides of the angle.
2. Converse theorem:
A point inside an angle and equidistant from both sides of the angle is on the bisector of the angle.
3. 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).
4. Use a ruler and compass to draw the bisector of the angle. (Method)
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