Miquel's Point of a 4-line Via Spiral Similarity

Spiral similarity is a combination of central similarity and rotation with the same center. Two circles are always similar and there is a continuum of spiral similarities that map one onto the other. Two finite straight line segments are also always similar. However, there exists exactly one spiral similarity that maps one onto the other. (Sometimes, spiral similarity reduces to one of its particular cases: rotation, central similarity, or translation. The latter happens wherever the two segments are equal and parallel. We shall overlook the exceptions.)

So, given two unequal and non parallel segments AB and CD. How to construct the center of the spiral similarity that rotates AB onto CD?

Here's the construction. Let P be the point of intersection of AB and CD. Besides P, the circumcircles of triangles ACP and BDP intersect at one other point, say, O. We claim that O is the sought center of spiral similarity. (This is also true if the two circles happen to be tangent to each other. Then P = O serves as the center of the spiral similarity.)

To prove the claim, suffice it to show that triangles AOB and COD are similar. This follows by comparing and equating inscribed angles. The actual derivation depends on the configuration. For example, depending on the configuration, we have either AOC + APC = 180^o and BPD + BOD = 180^o or AOC = APC and BPD = BOD. In both cases, APC = BPD. It thus follows that AOB = COD. Other angles may be considered in a similar fashion.

Assuming triangles AOB and COD are similar, we obtain OA/OC = OB/OD so that O is indeed the sought center.

This ratio entails an interesting bonus. Observe that triangles AOB and COD are similar iff triangles AOC and BOD are similar. Therefore, if AB is mapped on CD by a spiral similarity, so then AC is mapped on BD by a spiral similarity. This means that the point O is also the intersection of the circumcircles of triangles ABQ and CDQ, where Q is the point of intersection of AC and BD.

When both pairs AB and CD and AC and BD are extended as above, we may think of the configuration as consisting of four lines, a 4-line. The four lines taken three at a time form four triangles. What we have shown is that the circumcircles of the four triangles are concurrent at O, the center of spiral similarities, one taking AB onto CD, the other mapping AC onto BD.

O thus is Miquel's point of the 4-line.