The problem below is based on Hartshorne's Geometry: Euclid and Beyond. I request verification, critique, improvement, or feedback on my solution below.
Problem: Let two circles $r, s$ be tangent at point $A$. Draw two lines $EC$ and $DB$ through $A$, such that $E$ is on the top of $r$, $C$ on the bottom of $s$, $D$ on the bottom of $r$, and $B$ on top of $s$. Show that $BC \parallel DE$.
Solution: Draw the tangent line through $A$, labeling the top $P$ and the bottom $Q$. Observe that $\angle D \cong \angle EAP$, since they both subtend the same arc, and for the same reason, $\angle B \cong \angle CAQ$. But $\angle EAP$ and $\angle CAQ$ are vertical angles and therefore congruent, so $\angle D \cong \angle B$, and $DE \parallel BC$.
Discussion: This result can be generalized to: Given two tangent circles, the chords formed by lines which intersect at the point of tangency are parallel, subtend arcs of equal measure, and, when connected to the point of tangency, form similar triangles.
This generalization suggests an alternate approach, drawing the line $d$ through both centers and $A$ and noting equality of arcs between $E$ and $d$ and between $C$ and $d$. This approach has an appeal, in that it seems to work with elements more "fundamental" to the result. But I couldn't get it to be as simple as the above.
Questions:
- Is my solution correct?
- Can it, or its exposition, be improved?
- Is there a better approach? Thoughts on my alternate approach?
- Does this generalize in any interesting ways?
It generalizes to Reim's Theorem, a little known theorem which adds two things to the above:
While simple, the theorem is applicable to a range of cases whenever two circles intersect:
"Reim's theorem... every time you have intersecting circles, you know its there somewhere... e.g. Miquel, Mannheim, spiral similarities etc."
"The delight of the theorem is in its many converses and special cases, applicable to a wide range of geometric figures"