 The Department of Mathematics Education
J. Wilson, EMAT 6680

Using Geometer's Sketchpad to Develop Geometric Proofs

by: David Wise

Geometer's Sketchpad (GSP) is a powerful tool in helping students to discover geometric relationships, make conjectures, and develop proofs. However, emphasis must be placed on the fact that sketches are experiments that help support or refute, but not prove a conjecture. Students must learn to view sketches as an instrument that helps to demonstrate the underlying geometric relationships that can be used to prove the conjecture being investigated. Occassionally, a sketch can be constructed to develop a highly visual proof, but even in this case, a written explaination must accompany the sketch for the proof to be clear and complete. Particularly in high school geometry courses, the distinction between using a GSP sketch as an experiment and not the proof of the geometric concept is extremely important. This distinction must be made early and emphasized through the course. An important way for this to be accomplished, is for the teacher to not only use GSP for discovery demonstrations, but also to use sketches in the development of proof demonstrations. The following is a demonstration of how to use GSP in the development of a proof.

GSP is a dynamic geometric construction computer software package. For more information, contact the publisher, Key Curriculum Press. To see a GSP example, click here and then double-click on "Animate." To find instructions on setting up GSP as a helper application click here.

Theorem: The three perpendicular bisectors of a triangle are concurrent.

Given: Triangle ABX

Prove: The 3 perpendicular bisectors of triangle ABX are concurrent.

Proof: 1. Let D be the midpoint of segment AB, E be the midpoint of BX, and F be the midpoint of AX. Therefore, D bisects segment AB, E bisects BX, and F bisects AX. 2. Construct line m perpendicular to segment AB passing through point D. Therefore, line m is the perpendicular bisector of segment AB by definition.

3. Construct line n perpendicular to segment BX passing through point E. Therefore, line n is the perpendicular bisector of segment BX by definition.

4. Line m and line n intersect at point C because non-parallel lines intersect at one point. 5. Point C is equidistant from points A and B because a point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment. Therefore, segment AC = segment BC.

6. Point C is also equidistant from points B and X because a point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment. Therefore, segment BC = segment XC. 7. Since segment AC = segment BC and segment BC = segment XC, segment AC = segment XC through the substitution property of equality.

8. Therefore, point C is equidistant from points A and X, so point C must lie on the perpendicular bisector of segment AX because a point equidistant from the endpoints of a segment must lie on the perpendicular bisector of a segment.

9. Therefore, the 3 perpendicular bisectors of triangle ABX are concurrent. This point of concurrency is called the circumcenter.

Extension: Since segment AC = segment BC = segment XC, a circumscribed circle about triangle ABX can be easily constructed using point C as the center and segment AC, BC, or XC as the radius. This circle is called the circumcircle, and therefore, the point of concurrency created from the three perpendicular bisectors is called the circumcenter.

Click here to manipulate the sketch used throughout this proof to look at the circumcenter in relation to the type of triangle.

If you have any comments concerning this investigation that would be useful, especially for use at the high school level, please send e-mail to esiwdivad@yahoo.com.