Download Activity 5.2.2b Proof of the Perpendicular Bisector Theorem

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

History of trigonometry wikipedia , lookup

Euler angles wikipedia , lookup

Duality (projective geometry) wikipedia , lookup

Cartesian coordinate system wikipedia , lookup

Brouwer fixed-point theorem wikipedia , lookup

Pythagorean theorem wikipedia , lookup

History of geometry wikipedia , lookup

Four color theorem wikipedia , lookup

Line (geometry) wikipedia , lookup

Euclidean geometry wikipedia , lookup

Transcript
Name:
Date:
Page 1 of 2
Activity 5.2.2b Proof of the Perpendicular Bisector Theorem
The Perpendicular Bisector Theorems says that the locus of points that are equidistant from the
endpoints of a segment is the perpendicular bisector of the segment.
To prove this theorem we need to prove two things:
(1) If a point lies on the perpendicular bisector of a line segment, then it is equidistant from
the endpoints of the segment, and
(2) If a point is equidistant from the endpoints of a segment, then it lies on the perpendicular
bisector of the segment.
Prove part (1):
Given: ⃑𝑫π‘ͺ is the perpendicular bisector of Μ…Μ…Μ…Μ…
𝑨𝑩
⃑
P lies on 𝑫π‘ͺ
Prove: PA = PB
Fill in the blanks to complete the proof below
Statements
Reasons
⃑𝑫π‘ͺ is the perpendicular bisector of Μ…Μ…Μ…Μ…
𝑨𝑩
⃑
and P lies on 𝑫π‘ͺ
Given
AC = CB
(a) Definition of _________________________
βˆ π΄πΆπ‘ƒ and βˆ π΅πΆπ‘ƒ are both right angles
(b) Definition of _________________________
(c) _______________________________
All right angles are congruent
PC = PC
(d) ___________________________________
(e) _______________________________
(f) ___________________________________
(g) _______________________________
Corresponding parts of congruent triangles are
congruent
Activity 5.2.2b
Connecticut Core Geometry Curriculum Version 3.0
Name:
Date:
Page 2 of 2
Prove part (2)
Given: PA = PB
C is the midpoint of Μ…Μ…Μ…Μ…
𝐴𝐡
Prove: ⃑𝑃𝐢 βŠ₯ Μ…Μ…Μ…Μ…
𝐴𝐡
Statements
Reasons
PA = PB
Given
C is the midpoint of Μ…Μ…Μ…Μ…
𝑨𝑩
Given
AC = CB
(a) Definition of _________________________
(b) _______________________________
Reflexive Property
(c) _______________________________
(d) ___________________________________
π‘šβˆ π΄πΆπ‘ƒ = π‘šβˆ π΅πΆπ‘ƒ
(e) ___________________________________
(f) π‘šβˆ ____________ + π‘šβˆ __________ = 180°
(g) ___________________________________
π‘šβˆ π΄πΆπ‘ƒ + π‘šβˆ π΄πΆπ‘ƒ = 180°
Substitution (substitute π‘šβˆ π΄πΆπ‘ƒ for π‘šβˆ π΅πΆπ‘ƒ)
2π‘šβˆ π΄πΆπ‘ƒ = 180°
(h) __________________________________
(i) ______________________________
Division property of equality
(j) ______________________________
Definition of perpendicular lines
Activity 5.2.2b
Connecticut Core Geometry Curriculum Version 3.0