Transforming Geometric Proof with Reflections, Rotations and Translations

Transforming Geometric Proof with Reflections, Rotations and Translations Kristin A. Camenga, Houghton College NCTM 2011 Annual Meeting:– Sample proofs of workshop problems

Try to think how you would prove the following standard theorems using transformational proof!

1. Vertical angles are congruent.

Given: Two lines and which cross at point P. Prove:

Idea: Rotate 180⁰ about intersection point P.

Rotate the figure 180⁰ about point . Lines and will rotate to themselves since line is on both lines. In particular, ray rotates to ray and ray rotates to ray. Since these rays define the angles and angle rotates to so .

Alternate format (2 column) , using Rp,180(X) to note the rotation of object X around point 180 degrees and writing two geometric objects are equal if they coincide.

Statement Reason

1. Lines and which cross at . 1. Given

2. Rp,180()= 2. The image of a line about a point on the line

Rp,180()= by rotation of 180⁰ is the original line.

3. Rp,180()= 3. A rotation of 180 will cause points on a line

Rp,180()= through the center to switch sides of the point

of rotation.

4. Rp,180 = 4. The angles are composed of two rays which

rotate to each other.

5. 5. Reflections preserve angles.

2. If the base angles of a triangle are congruent, then the sides opposite those angles are congruent. Given: , where Prove: Idea: Reflect triangle over the perpendicular bisector of the base. Let be the midpoint of and define line m as the perpendicular bisector of . Reflect the figure over line m. Since ⊥ m, reflects to itself. Since is the midpoint, and reflects to and vice versa. Since the vertex and one side of reflect to the vertex and one side of and , the two angles reflect to each other and the ray reflects to ray and vice versa. We now have two lines, and which reflect to two lines, and , respectively. Two lines intersect at only one point, so , the intersection of and , reflects to itself, the intersection of and . Since reflects to itself and reflects to , reflects to and .

3. If a quadrilateral has diagonals that are perpendicular bisectors of each other, then it is a rhombus. Given: and are perpendicular bisectors of each other, with intersection . Prove: Quadrilateral is a rhombus

Rotate 180⁰ around point . and each rotate to themselves since lines rotated 180⁰ about a point on the line rotate to themselves. Since and are perpendicular bisectors of each other, is the midpoint of both, so and . Since they are the same distance from on a line that rotates to itself, and rotate to each other and and rotate to each other. Since two points determine a unique segment, segment rotates to segment and segment rotates to segment . Therefore and .

Reflect over line . and reflect to themselves since they are on the line of reflection. Since , reflects to itself. Since is on and , reflects to . Since two points determine a unique segment, segment reflects to segment and segment reflects to segment . Therefore and

Then by transitivity, , and is a rhombus.

4. SAS: If two right triangles have two corresponding pairs of sides congruent and the included angles congruent, then the triangles are congruent. Given: , , Prove Translate so that coincides with . Rotate so that coincides with . Since , coincides with . If and are on different sides of , reflect over Since and and coincide and are on the same side of the angle, coincides with. Since the angles coincide, the other rays and coincide. Since , coincides with . Since coincides with and with and two points determine exactly one line, coincides with . Since all the sides coincide, the angles do, as well. Since all sides and angles coincide, 5. HL: If two right triangles have congruent hypotenuses and one pair of legs congruent, then the triangles are congruent.

Given: and right angles; ; Prove:

Translate point to point . Rotate until coincides with . Since , coincides with . If and are on opposite sides of , reflect over . Since and are both right angles and right angles are congruent, the vertex and one side of the angles coincide, and the other side is on the same side of line , the angles coincide. Therefore the other sides coincide: ray coincides with ray . Now to use the other piece of the given information: Since , both and are on the circle with center and radius . On the other hand, and are on the same ray . Since the hypotenuse is longer than the leg, and are less than . This means is inside the circle with center and radius . Therefore, the ray starts inside the circle and continues in only one direction, so it only intersects the circle once. Therefore coincides with . Since all three vertices coincide, all sides coincide since two points determine a unique segment. Since all vertices and sides coincide, the triangles are congruent, so [Note: It is clear from this proof why it wouldn’t work if one of the sides is not the hypotenuse – and in fact, that SSA congruence works if the first side is the longer of the two pairs of sides!]