Scaling of Distances and Areas Under Homothety
Statement
Under a homothety with center and ratio , for any two points with images and , we have and hence . Consequently, the area of any triangle scales by the square of the ratio: .
Why is it true?
Because every point is pushed away from (or pulled toward) the center by the same factor , the triangle formed by and any two points is scaled uniformly in both radial sides, so by Thales' theorem the third side stays parallel to and scales by . Area is two-dimensional (base times height), and since both base and height are multiplied by , their product is multiplied by .
Proof sketch
By the definition of homothety, and . Subtracting the second vector equation from the first yields .
Using the head-to-tail rule on both sides, we obtain . Taking lengths of both vectors immediately gives .
Now consider any triangle . Its altitude from to line is the distance between and its orthogonal projection on . Since homothety preserves parallelism and angles, it preserves perpendicularity, so is the foot of the altitude of . Thus both the base and the altitude scale by , giving .
Topics that use this theorem
Step-by-step proofs
No step-by-step proof yet for this theorem.
References
- H. S. M. Coxeter (1969). Introduction to Geometry
- Wikipedia contributors (2026). Geometric transformation — Wikipedia