## Equal Chords in Crossing Circles

Here's a problem from an old geometry book by H. S. Hall and F. H. Stevens. The book is an augmented edition of Euclid's Elements, I - VI, with parts from book XI. For every Euclid's proposition, the authors he authors added a list of problems that illustrate or make use of that proposition. The problem at hand is listed as exercise #7 in the section 'Theorems and Examples on Book III', p. 234):

If two circles cut one another, any two straight lines drawn through a point of section, making equal angles with the common chord, and terminated by the circumferences, are equal.

(In the applet, the two circles are defined by three points each: P, Q, A, for one, and P, Q, D, for the other. Dragging either P or Q modifies the circles. Dragging either A or D may have different effect depending on which of the buttons at the bottom of the applet is checked. If it's "Adjust circles" then the circles will be modified. If the "Adjust chord" button is checked, A and B would be dragged over existing circles supplying a set of possible locations for the segments AB and CD.

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Solution

### References

1. H. S. Hall, F. H. Stevens, A text-book of Euclid's elements for the use of schools, MacMillan & Co, 1904

If two circles cut one another, any two straight lines drawn through a point of section, making equal angles with the common chord, and terminated by the circumferences, are are equal.

The problem is an easy consequence of another one:

If two circles cut one another, any two parallel straight lines drawn through the points of intersection to cut the circle, are equal.

To see how the two problems are related, draw through Q a line parallel to CD with a cutoff segment C'D'.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

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Then, by the previous problem, CD = C'D'. On the other hand, because of the equality of the angles with the common chord PQ, C'D' is symmetric with AB in the line of centers EF. So that C'D' = AB, implying AB = CD of course.

Hubert Shutrick found a simpler solution. As a matter of fact, both AB and CD equal twice the projection of EF on each. Since AB and CD are equally inclined relative to PQ whilst EF perpendicular to the latter, the two projections are equal and so are AB and CD.

### Acknowledgement

The remarkable book by H. S. Hall and F. H. Stevens was brought to my attention by Hubert Shutrick who quoted from his private correspondence with Gordon Walsh. The book is quite remarkable in its treatment of Euclid's Elements in part because it is was published at the time when most educators began to search for ways to exclude the Elements from school. H. S. Hall and F. H. Stevens interspersed Euclid's propositions with a multitude of exercises that help students internalize Euclid's logic and master his techniques. The book shows how the Elements could be incorporated into a school program. I would prefer it to many a textbook written in the century since.