The dilation with center O and ratio r is a transformation of the plane that sends a point P to the point P' so that P' is on line OP and OP' = r OP.
The figures in the previous section give examples of dilations. The dilation with center O and ratio OA1/OA dilates point B to point B1 and point C to point C1.
There is a built-in way to dilate points in Sketchpad, in fact more than one way.
In a new sketch, draw a triangle ABC and also a point O.
Select O and Mark O as Center (Transform Menu). Then Select the triangle ABC
and choose Dilate from the Transform Menu. You will have the opportunity to
type in a ratio. Type into the boxes the ratio 1 to 2. This will create
triangle A'B'C'.
Now drag the points ABC around to convince yourself the two
triangles are similar. But the relationship is stronger than that. Drag O so
that it is superimposed on one of the vertices, such as A. What is triangle
A'B'C' in this case?
Construct the segments lines OA,
Now repeat the same experiment, but this time type in the
ratio –1 to 2. Also construct the lines (not segments) OA,
Now drag O so that A'B'C' becomes the midpoint triangle. See how O is located at the centroid of the triangle.
In a new figure, draw a triangle ABC and the midpoints of its sides.
Draw a polygon, say a quadrilateral ABCD. Also draw a point O.
Now drag A' up and down the line so that you understand what ratio you have measured.
In particular, what is the ratio when A' is located at O, at
A, at the midpoint of OA. When is the ratio –1? When is the ratio 2?
Now, you can use the marked ratio to dilate the
quadrilateral. First, select O and choose Transform >
Move point A' so that O is between A and A'. What happens to the ratio? What happens to the dilated shape?
Special case:
Describe what happens when the ratio becomes -1.
Theorem 1: Let O be a point, r a ratio, and m a line. Then let A' and B' be the dilations of two points A and B by ratio r with center O. Then distance A'B' = r AB for any A or B.
Figure with proof. Construct a figure with points O, A, B, A'
and B' as described. Then note that
triangle OAB is similar to OA'B' with ratio r = OA'/OA =
Theorem 2: Let O be a point, r a ratio, and m a line. Then the dilation with center O by ratio r of all the points of m (i.e., the image of m) is a line m' parallel to m (except for the case then O is on m, when m' = m).
This experiment illustrates the point of the Theorem 2. It is not a proof.
The proof of the theorem follows from Thales theorem and SAS for similar triangles.
Make a figure like this.
In this figure you can say that DE is a dilation of BC with center A (and positive ratio r) or that ED is the dilation of BC with center F (and ratio –r). This means that AD/AB = AE/AC = DE/BC/ = r but FD/FC = FE/FB = ED/BC = -r.
Explain why the line AF intersects ED and BC in the midpoints of the segments.
In BG, it is explained how to divide a segment internally and externally in the ratio b to c. See BG pp 53-54. The figures look like this. You can do this in Sketchpad with one figure. Do you see the dilation that takes segment b to segment c? What is the center?