Appetizers and Lessons for Mathematics and Reason   
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1A. Pattern Based Reason  1995
1B. Math Curriculum Notes 1996
2. Three Skills for Algebra  1995
3.
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1. Arithmetic Videos  11-2008
2.  Algebra Videos (to appear)
3. Solving Linear Equations  04-2005
4.-Fractions-Rates-Proportns-Units-2006
5.  Algebra, Odds & Ends, HS level-2001
6.-Euclidean-Geometry/Complex No.s 
7.  Analytic Geometry/Functions 2006
8.  Number Theory. 2006-7
9.  Complex Numbers More 2001. 
10  Exponents & Radicals Exactly 2008
11. Calculus  2005

12.Real  Analysis 1995
13. Electric Circuits Etc  2007
Mathematics How TOs & site 
content guides  08- 2008
1. Arithmetic
2. Algebra 
3. More Algebra 
4. Geometry  
5. More Geometry
6. Calculus
Right Bisector Construction, Etc.        Back ] Home ] Next ]

20-August-2008

Construction of a Right Bisector of a Line Segment AB

Step 1.  Draw a circle with diameter 2r > the length of line segment AB about one of the end points:

Step 2: Draw  a second circle of same radius r about the other end point.

Step 3. Label the points of intersection of the two circles with letters P and Q.

Step 4.  Draw the line through the intersection points PQ or the line segment PQ only.

Step 5. Label the intersection of the line segment PQ and line segments AB with a letter D.

The construction is complete. 

Claim: The point D bisects the line segment AB and the line PQ is perpendicular to the line segment AB.  It or the line that extends provides a or the right bisector of the line segment AB.

Proof of the Claim:  Draw triangles PAB and QAB. These triangles are isoceles as both P and Q are at distance r from both A and B.

Moreover, triangles PAB and QAB are isometric by the SSS criteria. Therefore the four angles at A and B in triangles PAB and QAB are all equal:

Next, triangles APQ and BPQ are isometric by the SSS criteria, and they are also isosceles since adjacent sides at A and B in both are equal in length to the radius r. Therefore the four angles at P and Q in these triangles are equal.

Finally triangles AQD, APD, BPD and BQD are isometric by the ASA criteria. The latter isometry implies the four equal angles at D sum to 360 degrees (4 rights angles), and hence each must be a right angle . The latter isometry implies line segment AD and BD are equal in length. So the point D bisect the line segement AD.  And as a after thought, we observe that line segments PD and QD are of equal length.

Exercise: Show if  D bisects a line segment AB and CD is a line segment with  angles CDA and CDB forming  right angles than C is equidistant from A and B.  

Hint: Apply the SAS isometry criteria. 

The exercise show that each point C on a right bisector of line segment AB is equidistant from line segment endpoints A and B. The line extending  line segment PQ constructed above is a right bisector.  

 

Euclidean Geometry
with a geometry based
based development of 
complex numbers


24 Lessons:

Correspondence
Isometry
Side-Side-Side
Side Angle Side
Angle-Side-Angle
Isoceles
Right Bisector Construction, Etc.
Perpendicular - Point to Line
SSS Failure
SAS Failure
ASA Failure
Parallel Lines
Angle Sum
Similarity
Right Triangle Similarity
Trig  or Similarity
Parallelograms
Kites From Triangles Duplication
Parallelogram from Triangle Duplication
Addition of points in the plane
Multiplication of Points in the Plane
Distributive Law, Step I
Distributive Law, Step II
Distributive Law, Step III

Easy Consequences of  this (newest) Complex Number. Starter Lesson  in this site folder follow below.

Vec & Cmplx  No Applet
B2 C. Conjugates
B3 Pythagoras
B4 Distance
B5 Rt Triangle Similarity
B6 Trig., Functions
B7 Dot & Cross Products
B8 Cosine Law
B9 Exponential & cis fns
B10 Easy Trig Identities
B11 Set Viewpoint

 


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