Collinearity via Concyclicity
What Is This About?
A Mathematical Droodle

---

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 http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

---

Buy this applet What if applet does not run?

Explanation

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2012 Alexander Bogomolny

The applet below provides an illustration to a problem from an outstanding collection by T. Andreescu and R. Gelca:

Let ABC be a triangle and let D and E be points on the sides AB and AC, respectively, such that DE is parallel to BC. Let P be any point interior to triangle ADE and let F and G be the intersections of DE with the lines BP and CP, respectively. Let Q be the second intersection point of the circumcircles of triangles PDG and PFE. Prove that the points A, P, and Q lie on a straight line.

---

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 http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

---

If A, P, and Q are collinear, there are two pairs of secants to circles (secants ADM/APQ, circle DPG, and secants AEN/APQ, circle PFE). By the Intersecting Secants or Power of a Point theorems (and their inverses), to show that A, P, and Q are collinear, it would suffice to demonstrate that

AD × AM = AE × AN.

This is equivalent to proving that points D, M, E, N are concyclic. However, it is easier to prove that B, M, C, N are concyclic which would imply

AB × AM = AC × AN.

and then employ DE||BC, i.e., AD/AB = AE/AC, to conclude that AD × AM = AE × AN.

Points M, D, P, G are concyclic by the construction. Angles DMP and DGP are subtended by the same arc, so are either equal or supplementary. However, since DE||BC, ∠DGP = ∠BCP. Taking into account that angles DMP and BDP are supplementary, we see that points B, M, P, C are concyclic. Similarly, points B, N, P, C are also concyclic, so all five - B, C, M, N, P - are. As we already remarked, this implies that D, E, M, N are concyclic, and we are done.

Note that, as the solution shows, the requirement that P lies in the interior of ΔADE is rather spurious. The construction goes through for P anywhere in the plane, except lines BC and DE; the argument remains valid for all such P.

References

1. T. Andreescu, R. Gelca, Mathematical Olympiad Challenges, Birkhäuser, 2004, 5^th printing, 1.3.10 (p. 13)

Power of a Point wrt a Circle

1. Power of a Point Theorem
2. A Neglected Pythagorean-Like Formula
3. Collinearity with the Orthocenter
4. Circles On Cevians
5. Collinearity via Concyclicity
6. Altitudes and the Power of a Point
7. Three Points Casey's Theorem
8. Terquem's Theorem
9. Intersecting Chords Theorem

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2012 Alexander Bogomolny