Transformational Proof: Informal and Formal Kristin A. Camenga Kristin.camenga@houghton.edu Houghton College November 12,2009

A General Approach to Solving Problems Data Representation CLAIM Theorems Analysis

We used this approach to justify the claim that we could construct congruent angles. Representation FBE, rays j & l Data compass length BE, circle radius BE Analysis compass length constant = circles congruent Theorems Congruent circles = congruent radii SSS, CPCTC How did we know these we’re true?

What is a “proof”? In geometry, a proof is the justification of a statement or claim through deductive reasoning. Statements that are proven are called theorems. To complete the reasoning process, we use a variety of “tools” from our “toolbox”.

Toolbox Tools Given Information (data) Definitions often w/ respect to the representation Postulates (axioms) Properties (could be from algebra) Previously proved theorems Logic (analysis)

More visual and intuitive; dynamic In this unit, we will be doing informal transformational proofs. In the next unit we will be doing formal proofs. Why use this approach? More visual and intuitive; dynamic Helpful in understanding geometry historically In the proof of SAS congruence, Euclid writes “If the triangle ABC is superposed on the triangle DEF, and if the point A is placed on the point D and the straight line AB on DE, then the point B also coincides with E, because AB equals DE.” This is the idea of a transformation! Builds intuition and understanding of meaning Generalizes to other geometries more easily

Key ideas of Informal Transformational Proofs We already know these! Uses transformations: reflections, rotations, translations and compositions of these. Depends on properties of the transformation: Congruence is shown by showing one object is the image of the other under an isometry (preserves distance and angles)

Let’s talk through an example! link to… http://www.youtube.com/watch?v=O2XPy3ZLU7Y

Example: Show informally that the two triangles of a parallelogram formed by a diagonal are congruent. What do I have to write to “prove” this using transformations? Students can DISCOVER results and remember them this way. In fact we could DEFINE a parallelogram as a quadrilateral with 180 degree rotational symmetry around the midpoint of a diagonal.

Using patty paper, I can see that ∆ACD maps onto ∆DBA through a rotation. (Do you see that it’s not a reflection?) What is the nature of the rotation? Students can DISCOVER results and remember them this way. In fact we could DEFINE a parallelogram as a quadrilateral with 180 degree rotational symmetry around the midpoint of a diagonal. 10

To find the center of rotation, I connect two corresponding vertices To find the center of rotation, I connect two corresponding vertices. (∆ACD maps onto ∆DBA) I see that the center of rotation is point P. Is there anything special about P? What is the degree measure of the rotation? Students can DISCOVER results and remember them this way. In fact we could DEFINE a parallelogram as a quadrilateral with 180 degree rotational symmetry around the midpoint of a diagonal. Use patty paper and a protractor. 11

Informal Proof (what you have to write) ∆ACD maps onto ∆DBA by R(180◦ , P) where P is the midpoint of the diagonal. So ∆ACD ∆DBA Students can DISCOVER results and remember them this way. In fact we could DEFINE a parallelogram as a quadrilateral with 180 degree rotational symmetry around the midpoint of a diagonal. 12

Example: Parallelograms (Rigorous) Given: Parallelogram ABDC Draw diagonal AD and let P be the midpoint of AD. Rotate the figure 180⁰ about point P. Line AD rotates to itself. Since P is the midpoint of AD, PA≅PD and A and D rotate to each other. Since by definition of parallelogram, AB∥CD and AC∥BD, ∠BAD≅∠CDA and ∠CAD≅∠BDA. Therefore the two pairs of angles, ∠BAD and ∠CDA , and ∠CAD and ∠BDA, rotate to each other. Since the angles ∠CAD and ∠BDA coincide, the rays AC and DB coincide. Similarly, rays AB and DC coincide because ∠BAD and ∠CDA coincide. Since two lines intersect in only one point, C, the intersection of AC and DC, rotates to B, the intersection of DB and AB, and vice versa. Therefore the image of parallelogram ABDC is parallelogram DCAB. Based on what coincides, AC≅DB, AB≅DC, ∠B≅∠C, △ABD≅△DCA, and PC≅PB What CAN you prove? There are many results that can be remembered this way – and all gotten for the price of one! Notice that this uses another basic property of lines:

Today we are going to verify that isometries do preserve distance and angles. We call this Corresponding Parts (of) Congruent Figures (are)