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"Perpendicular bisectors of line segments" PPT courseware 8
teaching objectives
1. Be able to tell the theorem and converse theorem of the perpendicular bisector of a line segment, and be able to use these two theorems differently.
2. Understand the application of mathematics learning methods, observation, generalization, verification, comparison, etc. in this lesson.
3. Understand that mathematics comes from life and serves real life, and experience the application value of mathematics.
please think
1. How many isosceles triangles can be made with the known line segment AB as the base?
2. If you don’t use a ruler and compass but use a set square, can you draw the isosceles triangle required above?
3. If you only use a ruler, can you draw the isosceles triangle required above?
perpendicular bisector of line segment
Hands-on operation: Make a perpendicular bisect MN of line segment AB, and the vertical foot is C; pick any point P on MN to connect PA and PB; measure: the lengths of PA and PB, what can you find?
PA=PB P1A=P1B
What rules can you draw from this?
Proposition: The distance from a point on the perpendicular bisector of a line segment to both endpoints of the line segment is equal.
Property theorem: The distance from a point on the perpendicular bisector of a line segment to both endpoints of the line segment is equal.
Converse proposition: A point equidistant from both endpoints of a line segment is on the perpendicular bisector of the line segment.
Summary Expand
theorem
A point on the perpendicular bisector of a line segment is equidistant from both endpoints of the line segment.
As shown in the figure,
∵AC=BC,MN⊥AB,P is any point on MN (known),
∴PA=PB (the distance between the point on the perpendicular bisector of the line segment and the two endpoints of the line segment is equal).
Converse Theorem: A point equidistant from the two endpoints of a line segment lies on the perpendicular bisector of the line segment.
As shown in the figure,
∵PA=PB(known),
∴Point P is on the perpendicular bisector of AB (a point that is equidistant from the two endpoints of a line segment is on the perpendicular bisector of the line segment).
Example 1 Known: As shown in the figure, in ΔABC, the perpendicular bisectors of sides AB and BC intersect at P.
Verify: PA=PB=PC;
prove:
∵Point P is on the perpendicular bisector MN of line segment AB,
∴PA=PB(?).
In the same way PB=PC.
∴PA=PB=PC.
What conclusion can you draw from Example 1?
Conclusion: The perpendicular bisectors of the three sides of a triangle intersect at a point, and the distances from this point to the three vertices of the triangle are equal.
give it a try
Known: As shown in the figure, in the isosceles triangle ABC, the vertical horizontal line MN of the waist AB intersects AC at the point D, BC=8 cm,
The circumference of ΔBDC is 20 cm.
Find: the length of AB.
Known: As shown in the figure, D is a point on the extension line of BC, BD=BC+AC.
Prove: Point C is on the perpendicular bisector of AD.
summary
1. Property theorem: The distance from a point on the perpendicular bisector of a line segment to the two endpoints of the line segment is equal.
2. Converse theorem: A point equidistant from the two endpoints of a line segment is on the perpendicular bisector of the line segment.
Points on the perpendicular bisector of a line segment are equidistant from both endpoints of the line segment
A point equidistant from both endpoints of a line segment, on the perpendicular bisector of the line segment
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