I've been looking at spiral similarities as a problem solving tactic, which I have found surprisingly versatile (within synthetic geometry), however I have had trouble with a specific configuration.

I should mention, I've been using the following lemma to find this point:

Define $P$ to be the intersection of $AC$ and $BD$ (i.e. $P = AC\cap BD$). Circles $(ABP)$ and $(CDP)$ meet at $O$ ($O\neq P$); thus, $O$ is the center of the spiral similarity which takes $AB$ to $CD$.

But what do you do when one of the points ($A$, $B$, $C$ or $D$) is $P$? I know there should be a spiral similarity for them (because I heard there exists a unique spiral similarity for any two segments that are not parallel), but it is difficult because it means one of the circles above is only defined using two points, meaning I cannot fix the circle...

What do I need this for? There are problems where I think two triangles are similar, and I can't think of any other way of showing this than through spiral similarity. To do this I need to prove that some point is the spiral centre.

Thanks for any help you can provide.

  • $\begingroup$ Hint: Perpendicular at $P$ $\endgroup$ Commented Feb 4 at 4:49
  • $\begingroup$ Welcome to Math.SE! ... "But what do you do when one of the points ($A$, $B$, $C$ or $D$) is $P$?" ... Let's say $D=P$. Then the lemma's $\bigcirc ABP$ is well-defined, whereas $\bigcirc CDP=\bigcirc CPP$ is problematic. The trick here is to treat the double-pt as a pt of tangency with $BD$. (Imagine $D$ sliding toward $P$ along line $BD$.) The center is where the perpendicular to $BD$ at $D$ meets the perpendicular bisector of $CD$. Proving that $\triangle AOC\sim \triangle BOD$, as required for a spiral similarity with center $O$, is left as an exercise to the reader. :) $\endgroup$
    – Blue
    Commented Feb 4 at 4:51

1 Answer 1


I'm expanding a comment from Blue into an answer.

Without loss of generality, let's assume $P=D$ is the duplicate point. So you find that $AC$ intersects $BD$ in $D$. This makes $\bigcirc CDP$ undefined.

Now pick a different point $D'$ on line $BD$. Then $\bigcirc CD'P$ is well defined. You would construct its center for example by intersecting the perpendicular bisector of $CP$ with that of $D'P$.

Now we perform a limit process: $\lim_{D'\to D}$. You slide the point $D'$ towards $D$. In the limit, the two become the same. What happens to the perpendicular bisector of $D'P$ (or $D'D$ if you prefer)? It turns into the perpendicular to $BD$ in the point $D$. So the center and thus the circle can still be constructed in the limit.

Note that the outcome of the above process depends on the direction from which you approach the limit. I said you should pick $D'$ on line $BD$. But what would happen if you pick it on a different line? Then the perpendicular bisector of $D'P$ would be perpendicular to that other line, and in the limit would become the perpendicular to that other line. Fixing the line is therefore essential to making the outcome well-defined.

There is an important relation between the line and the resulting circle. The circle $\bigcirc CD'P$ intersects line $D'P$ in $D'$ and $P$. In the limit the two points of intersection become a single one with algebraic multiplicity two. That makes the line a tangent to the circle.

So by choosing $D'$ to lie on $BD$ we ensure that $BD$ is a tangent to $CD'P$ in the limit. That's important. In the general situation, with all points distinct, $\bigcirc CDP$ intersects $BD$ in $D$ and $P$. So in the special situation of $P=D$ we want that to become a single point of multiplicity two. You can think of that multiplicity as one accounting for $D$ and the other accounting for $P$. That's why we need $BD$ to be a tangent to the circle, and thus why I wanted $D'$ to lie on $BD$.


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