# Secant Angles in a Circle II

Secant is another name for a transversal. It's a straight line that crosses another shape of interest. Secants that cross a circle have many different properties. Below, our concern is with the angles formed by two secants that meet in a point, say, A. Such angles are appropriately called secant angles. In the applet, the lines can be dragged parallel to themselves or rotated around the points next to the applets' border. (The applet and the taken approach have been suggested by Stuart Anderson.)

With the exception of the case where A lies on the circle, the secant angle intercepts two arcs on the circle: BC and DE, in the applet. The secant angle equals half the sum of the angular measure of the two arcs (for A inside the circle) or half their difference (for A outside the circle). In the case where one of the secants is tangent to the circle, A still lies outside the circle and the angle intercepts two arcs. The angle is half their difference, as before.

### 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.

 What if applet does not run?

(A more traditional approach appears elsewhere.)

A key observation here is that we know up front that in some cases our statement is correct. Indeed, if the two lines cross at the center of the circle, they form a central angle whose measure is, by definition, the measure of the subtending arc. If we consider a pair of vertical angles, each of them will be equal to half the sum of the two arcs cut off by the straight lines.

Now, shift one of the lines so as to keep it parallel to its original position. This operation will not change the angle between the two lines. The arcs cut off by the two will however change. They will change by the same angular measure. Sometimes they will both grow or shrink and sometimes one of them will grow, while the other will decrease. The applet helps surmise when each of the cases takes place.

The important observation is that a parallel translation of a line does not violate the relationship between the angle and the cut-off arcs. The latter statement is true for small translations. To make it true universally, it is convenient to introduce the notion of a directed (or signed) arc.

To this end a direction (orientation) is chosen on the circle. It is either clockwise or anticlockwise. A circular arc is defined by two points of which is selected as the beginning and the other as the end of the arc. Such designation induces a direction on the arc: "from the beginning to the end". If this direction agrees with that of the circle, the arc's angular measure is considered positive. If the direction does not agree with the orientation of the circle, the arc's angular measure is considered negative.

In the context of the argument below, one of the chords is selected to be the "first", the other the "second". The arcs they form are measured from a point of intersection with the circle of the first line to the point of intersection of the second line. Thus, for example, if the two secants meet inside the circle, the two chords they form have the same signs. If the secants meet outside the circle, the two chords are of different signs.

In S. Anderson's words:

Let chords BD and CE cross at A (inside or outside the circle), and let them meet at a fixed angle theta. Then the sum of the (signed) arcs BC and DE is independent of the location of A.

Proof: Let B'D' be a new chord parallel to BD, meeting CE at A'. Then the (unsigned) arcs BB' and DD' are equal by symmetry around the common perpendicular bisector of BD and B'D'. But one of these arcs is added and one is removed from the arcs BC and DE to form arcs B'C and D'E. Therefore the sum of the arcs remains constant.

Similarly, one can move chord CE to a new position, causing A to slide along BD. The requirement that BD and CE be chords means that A must lie within a circumscribed parallelogram with sides parallel to BD and CE. Within this region, the two chord-sliding operations clearly suffice to move A from any point to any other point, therefore the sum of signed arcs is constant as required. QED

I specifically wanted to prove this without using the theorem that inscribed angles subtending the same arc are equal, because this is a simple consequence of the general result, and I wanted a "pure" proof. On the other hand, this would make a good example of a special case being equivalent to the general case.

General implies Special: Move A first to the center and second to the periphery of the circle. At the periphery, one arc vanishes, say, DE. Then arc BC is equal to the sum of the arcs found with A at the circle center, which are equal by symmetry. Therefore a given inscribed angle subtends twice the arc of the equal central angle.

Special implies General: Construct the extra line CD. Then angle DCE subtends arc DE and angle BDC subtends arc BC. Considering the triangle CAD and the two pairs of vertical angles at A shows that angle DCE + BDC = BAC = DAE. Therefore by the special case, an angle equal to BAC placed at the circle center would subtend an arc equal to half the sum of arcs BC and DE. Thus placing A at the circle center would cause two equal arcs to be subtended, each equal to half BC + DE. Therefore, the arcs subtended by intersecting chords at any location A sum to the same value as when A is at the circle center, and hence the sum of arcs BC and DE is constant.