"Perpendicular Bisector of Line Segment" Proof PPT Courseware

Description

"Perpendicular Bisector of Line Segment" Proof PPT Courseware

We once used origami to get:

A point on the perpendicular bisector of a line segment is equidistant from both endpoints of the line segment.

Can you prove this conclusion?

Known: As shown in the figure, AC=BC, MN⊥AB, P is any point on MN. Prove: PA=PB.

Analysis: (1) To prove that PA=PB, you need to prove that △APC≌△BPC where PA and PB are located,

The condition of △APC≌△BPC is given by

AC=BC,MN⊥AB, it can be inferred that it can satisfy the axiom (SAS).

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).

Summary and expansion

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).

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Update Time: 2024-11-21

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