# How to show parallelism

Problem: Given two non-congruent circles that intersect at two points X and Y. One secant segment passes through X and intersects one circle (C1) at A and the other circle (C2) at B. Another secant passes through Y and intersects C1 at C and C2 at D.

Show $\overline{AC}$$\parallel \overline{BD} • I have tried to show \angleABD and \angleDBA are supplementary. Here I get stuck because there is nothing to associated congruent angles between the different circles. The critical point can be shown that the measure of the arcs \stackrel\frown{XY} are different between the circles. • I have tried to show (extending \overline{BD} to \overleftrightarrow{B'D} with B'-B-D) \angleABD \cong \angleABB'. Same problem as above. • I have tried using the tangent lines at A, B, C, D to show the construction of a right angle from the intersection of the tangent lines at A and C with a segment through the center of C1 forms a right angle with \overline{AC}. Extending this segment to \overline{XY} creates a segment \perp with \overline{AC} at M. I can also do this with C2 to create a segment \perp with \overline{BD} at N. I then let the C1 segment intersect \overline{XY} at M' and the C2 segment \overline{XY} at N'. I have tried to show that \angleMM'Y \cong \angleNN'X which would show \overline{MM'}$$\parallel$ $\overline{NN'}$ which would lead me to both segments being $\perp$ to either $\overline{AC}$ or $\overline{BD}$. Thus both $\overline{AC}$ or $\overline{BD}$ would be $\perp$ to the same line and hence parallel. However, I have nothing that can be used to establish these two segments $\overline{MM'}$ and $\overline{NN'}$ are parallel. Distance to the circle centers is not established and angles between the segments with $\overline{XY}$ are not defined and not clearly derived.

The trouble I am having with all of these approaches is I am lacking some definite relationship such as two angles or segments being congruent that would bridge the gap so to speak to form an acceptable relationship. For instance, the circles are given as non-congruent. Thus, no relationship can be derived based on distance (that I can see) as the one reasonable value I can use - the radius - is different between the circles. The use of angle congruence is discouraged as the inscribed angles that may be derived are different between the circles due to the differing radii. There is no definite diameter that can be used to leverage right triangles as segments drawn through given points do not pass through the circle center. Furthermore, the right angles formed by the tangent lines focus all congruency at the segments $\overline{AC}$ and $\overline{BD}$. The segments $\perp$ from these congruencies at both circles do not form the same segment or line.

To state perhaps more generally, the luxury of have a congruency established based on symmetric relationships - as is frequent with these type of euclidian problems - is not apparent to me due to the non-congruent nature of the given circles.

I wish I was more talented to provide diagrams, but any insight to this - even a subtle hint - would be deeply appreciated.

• Hint: Opposite angles of a cyclic quadrilateral are supplementary. – Blue Feb 15 '15 at 17:57
• Brilliant. Thank you. – Chris Swanson Feb 15 '15 at 18:47

## 1 Answer

As I understand it, $\angle$CAX is supplementary to $\angle$CYX. $\angle$CYX forms a linear pair with $\angle$XYD, so $\angle$XYD is also supplementary to $\angle$CYX. By transitivity, $\angle$XYD $\cong$ $\angle$CAX. Furthermore, $\angle$XYD is supplementary to $\angle$XBD. Hence, $\angle$XBD is supplementary to $\angle$CAX. Therefore, $\overline{AC}$ $\parallel$ $\overline{BD}$ with interior angles on same side of transversal supplementary.

• Bingo! :) (FYI: It took me a bit to realize that this answer could be this straightforward.) – Blue Feb 15 '15 at 19:09