If two figures are similar and have the same orientation then there exists a homothecy that takes one of them into the other

Question

Theorem. If two figures are similar and have the same orientation then there exists a homothecy that takes one of them into the other.

I see this result is being used pretty often in problems involving homothecy, but I don't know how to prove it. I know the reciprocal is true. If we have two figures and there exists a homothecy that takes one into the other, then the figures are similar and have the same orientation. Please help me with a proof of the theorem. I am a beginner in homothecy.

Answer 1

Let's assume that the two figures aren't congruent (the "center" of the homothety will be at infinity).

Let's call the corresponding points on the two figures $A$ and $A'$. Firstly, let's prove the theorem is true for triangles.

Theorem: If two triangles $ABC$ and $A'B'C'$ are similar with ratio $r$ with the same orientation, then there is a homothety from one to the other.

Proof:

Let $P$ be the intersection of $AA'$ and $BB'$, and consider the homothety $\varphi$ sending $A\to A'$ and $B\to B'$ (because $AB||A'B',\triangle PAB\sim \triangle PA'B'$ so the ratios $$\frac{PA}{PA'}=\frac{PB}{PB'}$$ are equal, and thus such a homothety exists.

Let's consider where $\varphi(C)$ is. Because $\varphi$ sends $A\to A'$ and $AC||AC'$, we know that $\varphi(C)$ lies on ray $A'C'$. How far along this ray? Well, we know that $A'B'C'=rABC$ is the scaling up factor, and thus $C'$ lies at distance $rAC$ from $A'$.

However, we also know that $\varphi$ takes $AB$ to $A'B'$ so $\varphi$ also scales everything up by $r$. Hence, $$|\varphi(AC)|=r|AC|=|A'C'|$$ and thus $$C\xrightarrow{\;\;\;\varphi\;\;\;}C'\;\;\square$$

How does this help with the general case?

Well, consider two similar figures $\mathcal F$ and $\mathcal F'$. If each of these is a single point, then obviously there's a homothety between them.

Otherwise, fix any $A,B\in\mathcal F$, find the corresponding $A',B'\in\mathcal F'$, and consider defining $P,\varphi$ as before. Then for any other $C\in\mathcal F$, we know by the triangle theorem that $\varphi(C)=C'$, the corresponding point on $\mathcal F'$. Hence the two figures are related by the homothety $\varphi$ centered at $P$.

Answer 2

You can just join two pairs of corresponding points. The obtained lines meet at the center of homothety. If they are parallel, your center is located at infinity.